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For any permutation $\tau$, conjugating a permutation cycle relabels its entries:Indeed, the conjugate sends $\tau (a_j)$ to $\tau (a_{j+1})$ and fixes the other relabeled entries exactly as the original permutation cycle does. Therefore conjugation preserves the multiset of permutation cycle lengths. Conversely, if two permutations have the same cycle type, pair their permutation cycles of each length and let $\tau$ carry the $j$th entry of each first permutation cycle to the $j$th entry of its partner. Include one-cycles for fixed points. This defines a bijection of all labels, and the displayed relabeling identity makes the permutations conjugate. ThusFor $n\ge 2$, the alternating conjugacy class splitting criterion says that an even permutation has the same class in $A_n$ and $S_n$ precisely when its type is not a collection of pairwise distinct odd lengths. Equivalently, there is an even-length permutation cycle or a repeated permutation cycle length, counting each fixed point as a length-one permutation cycle. Distinct odd lengths instead give two alternating-group classes. For $n=1$, $A_1=S_1$ and the sole class is the same; the index-two splitting criterion is not applicable.

Squaring an odd-length permutation cycle keeps one permutation cycle of that length, since advancing by two visits all of its entries. Squaring an even-length permutation cycle splits it into two permutation cycles, each of half that length. Thus any even-length permutation cycle strictly increases the total permutation cycle count on squaring, while odd permutation cycles leave that count unchanged. The cycle type can agree with its square exactly when every length is odd. This proves permutation conjugate to its square:The final equivalence follows because the order of a group element that is a permutation is the least common multiple of its permutation cycle lengths.

The possible even types on five labels are the identity, a three-cycle, a product of two disjoint transpositions and a five-cycle. The identity and the double transpositions are their own inverses. For a three-cycle $(abc)$ with unused labels $d,e$, the even permutation $(bc)(de)$ conjugates it to its inverse. For a five-cycle $(abcde)$, the even permutation $(be)(cd)$ reverses its order. Therefore every element of $A_5$ is conjugate to its inverse within $A_5$. Inverting a five-cycle uses two transpositions, in agreement with inversion of an odd cycle in an alternating group.

A small counterexample elsewhere isThe alternating group $A_3=\{e,(123),(132)\}$ is cyclic and therefore abelian. Every conjugacy class in an abelian group is a singleton, while $\sigma \ne {\sigma }^{-1}$. Thus these two elements are not conjugate in $A_3$.