Inversion Mapping is Automorphism iff Group is Abelian

Theorem
Let $G$ be a group.

Let $\iota: G \to G$ be the inversion mapping on $G$, defined as:
 * $\forall g \in G: \iota \left({g}\right) = g^{-1}$

Then $\iota$ is an automorphism $G$ is abelian.

Proof
From Inversion Mapping is Permutation, $\iota$ is a permutation.

It remains to be shown that $\iota$ has the morphism property $G$ is abelian.

Sufficient Condition
Suppose $\iota$ is an automorphism.

Then:

Thus from Inverse of Commuting Pair, $x$ commutes with $y$.

This holds for all $x, y \in G$.

So $\left({G, \circ}\right)$ is abelian by definition.

Necessary Condition
Suppose $\left({G, \circ}\right)$ is abelian.

Thus $\iota$ has the morphism property and is therefore an automorphism.