The opposite-group functor is naturally isomorphic to the identity functor by inversion

Let $G^{\mathrm{op}}$ have the same set and identity as $G$, with $a\star b=ba$. A homomorphism $f:G\to H$ is also a homomorphism $G^{\mathrm{op}}\to H^{\mathrm{op}}$, since $f(a\star b)=f(b)f(a)=f(a)\star f(b)$. Thus $G\mapsto G^{\mathrm{op}}$ defines an endofunctor $O$ on $\mathbf{Grp}$.

Define $\nu_G:G\to G^{\mathrm{op}}$ by $\nu_G(a)=a^{-1}$. Then $\nu_G(ab)=b^{-1}a^{-1}=\nu_G(a)\star\nu_G(b)$, and $\nu_G$ is its own inverse as a set map, so it is a group isomorphism.

Every homomorphism preserves inverses, so for $f:G\to H$ one has $O(f)\nu_G(a)=f(a^{-1})=f(a)^{-1}=\nu_Hf(a)$. Hence the component square commutes.

The isomorphisms $\nu_G$ are natural by step 2.1. Therefore inversion defines a natural isomorphism $1_{\mathbf{Grp}}\cong O$.