= Solution
For any <permutation> $\tau$, conjugating a <permutation cycle> relabels its entries:
$$
\tau(a_1\,a_2\,\ldots,a_r)\tau^{-1}=(\tau(a_1)\,\tau(a_2)\,\ldots,\tau(a_r)).
$$
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. Thus
$$
\boxed{\sigma\sim_{S_n}\rho\quad\Longleftrightarrow\quad\sigma,\rho\text{ have the same cycle type}.}
$$
For $n\geq2$, 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, \b[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>:
$$
\boxed{\sigma\sim_{S_n}\sigma^2\quad\Longleftrightarrow\quad\text{every cycle length is odd}
\quad\Longleftrightarrow\quad\operatorname{ord}(\sigma)\text{ is odd}.}
$$
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 $(a\,b\,c)$ with unused labels $d,e$, the even <permutation> $(b\,c)(d\,e)$ conjugates it to its inverse. For a five-cycle $(a\,b\,c\,d\,e)$, the even <permutation> $(b\,e)(c\,d)$ reverses its order. Therefore \b[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 is
$$
\boxed{n=3,\qquad\sigma=(1\,2\,3).}
$$
The <alternating group> $A_3=\{e,(1\,2\,3),(1\,3\,2)\}$ 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 $A_3$.
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