Vrftsjtryk

Let $f$ and $g$ be permutations such that

$$f circ f = id,$$

$$g circ g = id,$$

and

$$fcirc g =begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \
10 & 4 & 5 & 7 & 8 & 9 & 2 & 6 & 3 & 1
end{pmatrix}.$$

Find $f$ and $g$.

I can solve it by a lot of guess work, but I wonder if there is some general method.

A permutation $f$ is an involution if $fcirc f=id$.

As you know, any permutation can be written as a product of disjoint cycles; your permutation is $(1 10)(2 4 7)(3 5 8 6 9)$. In order to write an arbitrary permutation as a product of two involutions, it suffices (since disjoint permutations commute) to write a cycle of arbitrary length as a product of two involutions. A permutation is an involution if it's a product of disjoint cycles of length $2$. If you experiment a little with multiplying involutions, you might discover the following pattern:
$$(1 2)circ(2 3)=(1 2 3)tag3$$
$$(1 2)(3 4)circ(2 3)=(1 2 4 3)tag4$$
$$(1 2)(3 4)circ(2 3)(4 5)=(1 2 4 5 3)tag5$$
$$(1 2)(3 4)(5 6)circ(2 3)(4 5)=(1 2 4 6 5 3)tag6$$
etc. So a cycle of any length can be obtained by multiplying two involutions. In particular, replacing $1,2,3$ by $2,4,7$ in $(3)$ we get
$$(2 4 7)=(2 4)circ(4 7);$$
replacing $1,2,4,5,3$ by $3,5,8,6,9$ in $(5)$ we get
$$(3 5 8 6 9)=(3 5)(9 8)circ(5 9)(8 6)=(3 5)(8 9)circ(5 9)(6 8);$$
and of course
$$(1 10)=(1 10)circ id;$$
so
$$(1 10)(2 4 7)(3 5 8 6 9)=(1 10)(2 4)(3 5)(8 9)circ(4 7)(5 9)(6 8).$$

I.e., you can take
$$f=(1 10)(2 4)(3 5)(8 9), g=(4 7)(5 9)(6 8).$$
Of course there are other solutions.

There is a well-known algorithm for decomposing any given permutation as a product of (not necessarily disjoint) $2$-cycles/transpositions. Such a decomposition of a given $fcirc g$ would, in general, give strong hints about (if not completely determine) the nature of $f$ and $g$.

Why?

Because here $f,g$ are involutions: they square to the identity. Thus they're each (either trivial or) determined by a product of disjoint $2$-cycles (a.k.a. transpositions) (although they may share one or more of those cycles), which are themselves involutions.

See @bof's answer for the same approach, fleshed out.

Here is how I did it. First write $fcirc g$ as a product of disjoint cycles. So here $fcirc g=(1, 10)(2,4,7)(3,5,8,6,9)$. Now

$$begin{align}
gcirc f &=fcirc(fcirc g)circ f \
&=fcirc(fcirc g)circ f^{-1} \
&=(f(1),f(10))(f(2), f(4),f(7))(f(3),f(5),f(8),f(6),f(9)).
end{align}$$

On the other hand $(fcirc g)^{-1}=g^{-1}circ f^{-1}=gcirc f$. So from the provided $fcirc g$ we can invert it to get $gcirc f$ which is $$(1,10)(2,7,4)(3,9,6,8,5)$$

Therefore $$(f(1),f(10))(f(2), f(4),f(7))(f(3),f(5),f(8),f(6),f(9))=(1,10)(2,7,4)(3,9,6,8,5)$$ and we can define $f$ to satisfy this (notice $f$ is not unique). After this you can find your $g$.