Characterize the groups $G$ for which the map $\iota: G \to G$, sending $x \mapsto x^{-1}$ for all $x \in G$, is an automorphism of $G$

I have trouble on this question: "Characterize the groups G for which the map $$\iota: G \to G$$ sending $$x \mapsto x^{-1}$$ for all $x \in G$, is an automorphism of $G$."

From this post I have seen that $G$ must be abelian. Can anyone propose another characteristic of $G$ that I can derive from the given conditions besides abelian? Thank you so much!!

Notice that $f(x)=x^{-1}$ is a one to one and onto function on $G$ since every group element has a unique inverse.

Thus if $f(xy)=f(x)f(y)$ for every pair in $G$ then it is an isomorphism from $G$ to $G$. But that means that $y^{-1}x^{-1}=x^{-1}y^{-1}$ $\implies$ $xy=yx$ for all pair in $G$ thus, $G$ is abelian.

Conversly if $G$ is abelian you can trivially show it is an isomorphism. Since it is an "if and only if condition", it is an "necessary and sufficient" condition. (you don't need to search for another condition)