Alternating group implies every element is automorphic to its inverse

Statement
Let $$n$$ be any natural number. Let $$A_n$$ denote the fact about::alternating group of degree $$n$$, i.e., the group of even permutations on a set of size $$n$$. Then, $$A_n$$ is a fact about::group in which every element is automorphic to its inverse. In other words, for every element $$g \in A_n$$, there is an automorphism of $$A_n$$ that conjugates $$g$$ to $$g^{-1}$$.

Related facts about alternating groups

 * Alternating group implies any two elements generating the same cyclic subgroup are automorphic
 * Classification of ambivalent alternating groups: $$A_n$$ is ambivalent for precisely the values $$n = 1,2,5,6,10,14$$.
 * Classification of alternating groups having a class-inverting automorphism: $$A_n$$ has a class-inverting automorphism for precisely the values $$n = 1,2,3,4,5,6,7,8,10,12,14$$.

Related facts about every element being automorphic to its inverse

 * General linear group implies every element is automorphic to its inverse
 * Projective general linear group implies every element is automorphic to its inverse
 * Special linear group implies every element is automorphic to its inverse
 * Projective special linear group implies every element is automorphic to its inverse
 * Every element is automorphic to its inverse is characteristic subgroup-closed
 * Normal subgroup of ambivalent group implies every element is automorphic to its inverse

Facts used

 * 1) uses::Symmetric groups are rational, or uses::cycle type determines conjugacy class

Proof
Consider the group $$S_n$$, in which $$A_n$$ is a normal subgroup of index two. Since $$S_n$$ is a rational group, the elements $$g$$ and $$g^{-1}$$ are conjugate. Thus, there exists $$x \in S_n$$ such that $$xgx^{-1} = g^{-1}$$.

Since $$A_n$$ is a normal subgroup of $$S_n$$, conjugation by $$x$$ restricts to an automorphism of $$A_n$$.