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When I was trying to explain a combinatorial curiosity using permutation groups, I finally ended up with another curiosity about the alternating group $A_5$. For any permutation $pi in S_5$, let $chi(pi)$ denote the number of fixed points of $pi$. Furthermore, as usual, let $[pi,sigma] := pi^{-1}sigma^{-1}pisigma$ denote the commutator of two permutations $pi, sigma$. Now I found out that the following holds:

Let $sigma in S_5$ be an arbitrary cycle of length 5. Then we have $$A_5 = { pi in S_5 : chi([sigma, pi]) + chi(sigma^2 [sigma, pi]) in {2,5} }. $$

Isn't that strange!? I verified that equation with a computer, but I have absolutely no idea how to prove it. Even proving the original combinatorial problem wouldn't help immediately, because the group theoretical statement is stronger.

Is there any chance to prove that characterization of $A_5$ by group theory?

To sketch a partial solution to this we puzzle we can use some group character theory. Below I have included the character tables of $Sym(5)$ and $Alt(5)$.
We first observe that the function

$$chi(g)=|fix(g)|$$

is a group character, and by a well known theorem (Theorem 2.6.1) we see that $chi(g)=rho_1(g)+rho_4(g)$, restricting to $Alt(5)$ we see that $chi(g)=phi_1(g)+phi_4(g)$.

Now fix $sigma=(12345)$ and pick and conjugacy class representatives in $Alt(5)$:

$pi_1=id$,

$pi_2=(12)(34)$,

$pi_3=sigma$

and $pi_4=sigma^{-1}$.

We have:

$[sigma,pi_1]=id$,

$[sigma,pi_2]=(12453)$,

$[sigma,pi_3]=(134)$,

$[sigma,pi_4]=id$

and $[sigma,pi_5]=(15432)=sigma^{-1}$.

Moreover,

$sigma^2[sigma,pi_1]=sigma^2$,

$sigma^2[sigma,pi_2]=(142)$,

$sigma^2[sigma,pi_3]=(15243)$,

$sigma^2[sigma,pi_4]=sigma^2$

and $sigma^2[sigma,pi_5]=(12345)=sigma$.

Now, evaluating the character

$phi_1([sigma,pi_i])+phi_4([sigma,pi_i])+phi_1(sigma^2[sigma,pi_i])+phi_4(sigma^2[sigma,pi_i])=chi([sigma,pi_i])+chi(sigma^2[sigma,pi_i])$

on those conjugacy classes gives either $2$ or $5$, and we're done.

For example, for $pi_2$ we have

$chi(sigma^2[sigma,pi_i])=phi_1([sigma,pi_2])+phi_4([sigma,pi_2])+phi_1(sigma^2[sigma,pi_2])+phi_4(sigma^2[sigma,pi_2])=phi_1((12453))+phi_4((12453))+phi_1((142))+phi_4((142))=1-1+1+1=2$.

-edit-

I'll include some remarks about why this not a complete solution and what remains to be proven. The key mystery seems to be that to obtain $2$ or $5$ we require the elements $[sigma,pi]$ and $sigma^2[sigma,pi]$ to have orders $(1,5)$, $(1,3)$, $(5,1)$, $(3,1)$ or $(2,2)$. Why this happens, I do not know. That being said, the fact that the commutator of two elements of $Alt(5)$ is clearly going to be contained in $Alt(5)$, also multiplying a commutator of $Alt(5)$ elements by an element of $Alt(5)$ will also be in $Alt(5)$.

The character table of $Sym(5)$:

$$
begin{array}{c|rrrrrrr}
rm class&rm1&rm2&rm2^2&rm3^1&rm4^1&rm5^1&rm6^1cr
rm size&1&10&15&20&30&24&20cr
hline
rho_{1}&1&1&1&1&1&1&1cr
rho_{2}&1&-1&1&1&-1&1&-1cr
rho_{3}&4&-2&0&1&0&-1&1cr
rho_{4}&4&2&0&1&0&-1&-1cr
rho_{5}&5&1&1&-1&-1&0&1cr
rho_{6}&5&-1&1&-1&1&0&-1cr
rho_{7}&6&0&-2&0&0&1&0cr
end{array}
$$

The character table of $Alt(5)$:

$$
begin{array}{c|rrrrr}
rm class&rm1&rm2^2&rm3^1&rm5_A&rm5_Bcr
rm size&1&15&20&12&12cr
hline
phi_{1}&1&1&1&1&1cr
phi_{2}&3&-1&0&frac{1+sqrt{5}}{2}&frac{1-sqrt{5}}{2}cr
phi_{3}&3&-1&0&frac{1-sqrt{5}}{2}&frac{1+sqrt{5}}{2}cr
phi_{4}&4&0&1&-1&-1cr
phi_{5}&5&1&-1&0&0cr
end{array}
$$