05AB1E, 5 bytes

ΣêR}R

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or as a Test suite

Explanation

Σ  }    # sort input by

 ê      # its sorted unique characters

  R     # reversed (to sort descending)

    R   # reverse the result (to sort descending)

Python 2, 60 55 bytes

def f(l):l.sort(None,lambda n:sorted(set(`n`))[::-1],1)

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Ungolfed

def f(l):

  l.sort(        # Sort the list in place

    cmp = None,  # ... with no custom comparison function

    key = k,     # ... on the function

    reverse = 1  # ... in reverse

  )              # As the arguments are used in the right order, no names are necessary.



k = lambda n:sorted( # sort  

  set(`n`)           # ... the set of digits

  )[::-1]            # reverse the result

                     # As '-' is smaller than the digits,

                     # it will be sorted to the back and ignored for sorting

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Perl 6, 36 34 33 31 bytes

-1 byte thanks to Jo King
-2 bytes thanks to Phil H

*.sort:{sort 1,|set -<<m:g/d/}

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Explanation

       {                      }  # Map each number, e.g. -373

                       m:g/d/  # Extract digits: (3, 7, 3)

                    -<<  # Negate each digit: (-3, -7, -3)

                set  # Convert to set to remove duplicates

               |  # Pass as list of pairs: (-3 => True, -7 => True)

             1,  # Prepend 1 for "none": (1, -3 => True, -7 => True)

        sort  # Sort (compares 1 and pair by string value): (-7 => True, -3 => True, 1)

*.sort:  # Sort lexicographically

Python 2, 58 60 bytes

lambda a:sorted(a,key=lambda x:sorted(set(`x`))[::-1])[::-1]


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Pyth, 7 6 bytes

-1 byte by @Sok

_o_{S`

Pyth, which uses only printable ASCII, is at a bit of a disadvantage here. Optimally encoded this would be 6*log(95)/log(256) = 4.927 bytes, beating 05AB1E.

Explained:

 o              Sort the implicit input by lambda N:

  _               reversed

   {               uniquified

    S               sorted

     '               string representation [of N]

_               then reverse the result.

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R, 97 95 bytes

function(x)x[rev(order(sapply(Map(sort,Map(unique,strsplit(paste(x),"")),T),Reduce,f=paste0)))]

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This challenge seems to have been pessimized for R. Explanation of the original version (start at 1. and work up):

f <- function(x) {

  x[                                                  # 8. Input vector in

    rev(                                              # 7. Reversed

        order(                                        # 6. Lexicographical order

              sapply(                                 # 5. Paste....

                     Map(sort,                        # 4. Sort each using...

                              Map(unique,             # 3. Deduplicate each

                                  strsplit(           # 2. Split each string into characters

                                           paste(x),  # 1. Coerce each number to string

                                           "")),      

                         T),                          # 4. ...descending sort.

                     paste,collapse="")               # 5. ...back into strings

              )

        )

    ]

}

Jelly, 8 bytes

ADṢUQµÞU

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How it works

ADṢUQµÞU  Main link (monad). Input: integer list

     µÞU  Sort by (reversed):

AD        Absolute value converted to decimal digits

  ṢUQ     Sort, reverse, take unique values

Brachylog, 9 bytes

{ȧdṫo₁}ᵒ¹

Note: due to how ordering works in brachylog, it does not work on number correctly. This is fixed by casting the number to a string (ṫ) at the cost of 1 byte.

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Stax, 6 7 bytes

èó≥ü≤♥¥

Run and debug it

MathGolf, 7 6 bytes

áÉ░▀zx

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Explanation

After looking at Emigna's 05AB1E solution, I found that I didn't need the absolute operator (and my previous answer was actually incorrect because of that operator). Now the main difference is that I convert to string and get unique characters instead of using the 1-byte operator in 05AB1E.

áÉ      Sort by the value generated from mapping each element using the next 3 instructions

  ░     Convert to string

   ▀    Get unique characters

    z   Sort reversed (last instruction of block)

     x  Reverse list (needed because I don't have a sort-reversed by mapping)

J, 17 bytes

{~[::~.@:~@":@|

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Explanation:

                @|    - find the absolute value and

             @":      - convert to string and

         @:~         - sort down and

       ~.             - keep only the unique symbols

    :                - grade down the entire list of strings   

  [:                  - function composition

{~                    - use the graded-down list to index into the input   

JavaScript (SpiderMonkey), 68 bytes

Thanks for @Arnauld for reminding me again that SpiderMonkey uses stable sort, so -4 bytes for removing ||-1.

A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y))

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JavaScript (Node.js), 72 bytes

A=>A.sort((x,y,F=n=>[...new Set(""+n)].sort().reverse())=>F(x)<F(y)||-1)

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Java (JDK), 98 bytes

l->l.sort((a,b)->{int r=0,i=58;for(;r==0&i-->48;)r=(b.indexOf(i)>>9)-(a.indexOf(i)>>9);return r;})

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Explanation

l->                           // Consumer<List<String>>

 l.sort(                      //  Use the incorporated sort method which uses a...

  (a,b)->{                    //   Comparator of Strings

   int r=0,                   //    define r as the result, initiated to 0

       i=58;                  //           i as the codepoint to test for.

   for(;r==0&i-->48;)         //    for each digit codepoint from '9' to '0',

                              //     and while no difference was found.

    r=                        //     set r as the difference between

     (b.indexOf(i)>>9)-       //      was the digit found in b? then 0 else -1 using the bit-shift operator

     (a.indexOf(i)>>9);       //      and was the digit found in a? then 0 else -1.

   return r;                  //    return the comparison result.

  }

 )

Note:

I needed a way to map numbers to either 0/1 or 0/-1.

indexOf has the nice property that it's consistently returning -1 for characters not found. -1 right-shifted by any number is always -1. Any positive number right-shifted by a big enough number will always produce 0.

So here we are:

input        input.indexOf('9')      input.indexOf('9')>>9

"999"        0                       0

"111119"     5                       0

"123456"     -1                      -1

C (gcc), 114 111 bytes

a;v(n){n=n<0?-n:n;for(a=0;n;n/=10)a|=1<<n%10;n=a;}c(int*a,int*b){a=v(*a)<v(*b);}f(a,n)int*a,n;{qsort(a,n,4,c);}

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Explanation:

f() uses qsort() to sort provided array in place. Using comparison function c() to compare numbers which evaluates numbers using v().
v() calculates a higher number if bigger digits are present in parameter.

[Edit 1]
Improved by 3 bytes. 2 byte credits to Kevin. Thanks

Japt, 12 bytes

ñ_a ì â ñnÃw

All test cases

Explanation:

ñ_        Ãw    :Sort Descending by:

  a             : Get the absolute value

    ì           : Get the digits

      â         : Remove duplicates

        ñn      : Sort the digits in descending order

Haskell, 54 52 bytes

import Data.List

f=r.sortOn(r.sort.nub.show);r=reverse

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Ruby, 55 bytes

->a{a.sort_by{|x|x.abs.digits.sort.reverse|}.reverse}

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Perl 5 -nl, 68 bytes

$a{eval"9876543210=~y/$_//dcr"}=$_}{say$a{$_}for reverse sort keys%a

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C# (Visual C# Interactive Compiler), 84 bytes

x=>x.OrderByDescending(y=>String.Concat((y+"-").Distinct().OrderByDescending(z=>z)))

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In C#, strings are considered "enumerables" of characters. I use this to my advantage by first converting each number to a string. LINQ is then leveraged to get the unique characters (digits) sorted in reverse order. I convert each sorted character array back into a string and use that as the sort key to order the whole list.

I was scratching my head with how to handle the hyphen for negative numbers. But as it turns out, adding the hyphen character to all of my numbers takes care of the tie breaker when one number has a superset of digits compared to the other.

Julia 1.0, 50 bytes

x->sort(x,by=y->sort(unique([digits(-abs(y));1])))

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APL (Dyalog Extended), 19 bytes

{⍵[⍒∪¨(∨'¯'~⍨⍕)¨⍵]}

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Fixed at a cost of +2 bytes thanks to OP.

APL(NARS), 366 chars, 732 bytes

_gb←⍬



∇a _s w;t

t←_gb[a]⋄_gb[a]←_gb[w]⋄_gb[w]←t

∇



∇(_f _q)w;l;r;ls;i

(l r)←w⋄→0×⍳l≥r⋄l _s⌊2÷⍨l+r⋄ls←i←l⋄→3

  →3×⍳∼0<_gb[i]_f _gb[l]⋄ls+←1⋄ls _s i

  →2×⍳r≥i+←1

l _s ls⋄_f _q l(ls-1)⋄_f _q(ls+1)r

∇



∇r←(a qsort)w

r←¯1⋄→0×⍳1≠⍴⍴w⋄_gb←w⋄a _q 1(↑⍴w)⋄r←_gb

∇



f←{∪t[⍒t←⍎¨⍕∣⍵]}



∇r←a c b;x;y;i;m

x←f a⋄y←f b⋄r←i←0⋄m←(↑⍴x)⌊(↑⍴y)⋄→3

→0×⍳x[i]<y[i]⋄→3×⍳∼x[i]>y[i]⋄r←1⋄→0

→2×⍳m≥i+←1⋄r←(↑⍴x)>(↑⍴y)

∇

For the qsort operator, it is one traslation in APL of algo page 139 K&R Linguaggio C.
I think in it there is using data as C with pointers...
Test

 c qsort 123, 478, ¯904, 62778, 0, ¯73, 8491, 3120, 6458, ¯7738, 373 

8491 ¯904 62778 478 ¯7738 6458 ¯73 373 3120 123 0 

 c qsort 11, ¯312, 902, 23, 321, 2132, 34202, ¯34, ¯382 

902 ¯382 34202 ¯34 321 ¯312 2132 23 11 

 c qsort 9, 44, 2212, 4, 6, 6, 1, 2, 192, 21, 29384, 0 

29384 192 9 6 6 4 44 2212 21 2 1 0 

 c qsort 44, ¯88, 9, 233, ¯3, 14, 101, 77, 555, 67 

9 ¯88 67 77 555 14 44 233 ¯3 101