Let $n\ge 3$, $\sigma\in A_n$, $cl_{S_n}(\sigma)$ be the conjugacy class of $\sigma$ in $S_n$ and $cl_{A_n}(\sigma)$ be the conjugacy class of $\sigma$ in $A_n$.
Show that if $C_{S_n}(\sigma)\subset A_n$ then $cl_{S_n}(\sigma)=cl_{A_n}(\sigma)\coprod cl_{A_n}(\tilde{\sigma})$ where $\tilde{\sigma}=\tau \sigma \tau^{-1}$ with $\tau\in S_n\backslash A_n$
Thanks for your help
First, note that $C_{A_n}(\sigma)=C_{S_n}(\sigma)\cap A_n$. In particular, if $C_{S_n}(\sigma)\subset A_n$, then $C_{S_n}(\sigma)=C_{A_n}(\sigma)$. Hence, $$|cl_{S_n}(\sigma)|=[S_n:C_{S_n}(\sigma)]=[S_n:C_{A_n}(\sigma)]=[S_n:A_n][A_n:C_{A_n}(\sigma)]=2|cl_{A_n}(\sigma)|.$$
Continue to assume $C_{S_n}(\sigma)\subset A_n$, and let $\tau\in S_n$ be any odd permutation. Observe that $\tau\sigma\tau^{-1}\notin cl_{A_n}(\sigma)$. Indeed, if there exists $\rho\in A_n$ such that $\rho\sigma\rho^{-1}=\tau\sigma\tau^{-1}$, then $\tau^{-1}\rho\in C_{S_n}(\sigma)\leq A_n$, a contradiction.
As distinct conjugacy classes are disjoint, it follows that $cl_{A_n}(\sigma)\sqcup cl_{A_n}(\tau\sigma\tau^{-1})\subset cl_{S_n}(\sigma)$. To show equality, it is enough to show $|cl_{A_n}(\sigma)|=|cl_{A_n}(\tau\sigma\tau^{-1})|$, since $|cl_{A_n}(\sigma)|=\frac{1}{2}|cl_{S_n}(\sigma)|$ by the first paragraph. But, this follows from the equality $\tau C_{A_n}(\sigma)\tau^{-1}=C_{A_n}(\tau\sigma\tau^{-1})$.
Indeed, $\tau\rho\tau^{-1}\in C_{A_n}(\tau\sigma\tau^{-1})$ whenever $\rho\in C_{A_n}(\sigma)$. Hence, $C_{A_n}(\tau\sigma\tau^{-1})\subseteq \tau C_{A_n}(\sigma)\tau^{-1}$ Now, conjugation is order preserving, so equality holds. Hence, $$ |cl_{A_n}(\sigma)|=[A_n:C_{A_n}(\sigma)]=[A_n:C_{A_n}(\tau\sigma\tau^{-1})]=|cl_{A_n}(\tau\sigma\tau^{-1})| $$ which completes the proof.