Inversion Mapping is Isomorphism to Opposite Group

Definition
Let $\left({G, \circ}\right)$ be a group.

Let $\left({G, *}\right)$ be its opposite group.

That is, for all $g, h \in G: g \circ h = h * g$.

Let $\iota: G \to G$ be the inversion mapping for $\left({G, \circ}\right)$.

Then $\iota: \left({G, \circ}\right) \to \left({G, *}\right)$ is a group isomorphism.

Proof
By Inversion Mapping is Involution, $\iota$ is an involution.

By Involution is Permutation, $\iota$ is a permutation and hence by definition a bijection.

It remains to show that $\iota$ is a group homomorphism.

Let $g, h \in G$.

Then:

Hence $\iota$ is a group homomorphism.

By definition it follows that $\iota$ is also a group isomorphism.