What are the auto-equivalences of the category of groups?
Just to make this a little more visible: for a proof that every autoequivalence of the category of groups is naturally isomorphic to the identity, see page 31 of Peter Freyd's book "Abelian Categories".
Your functor –op is naturally isomorphic to the identity, via the natural transformation G → Gop sending g to g-1. (If you asked about monoids, not groups, then –op would be a nontrivial autoequivalence.)
I'm pretty sure the category of groups has no nontrivial autoequivalences—more precisely the groupoid of autoequivalences is contractible. Certainly the identity autoequivalence has no natural automorphisms, because the only possible components of a natural automorphism on the object Z are +1 and -1, and by naturality the latter would have to extent to the "automorphism" sending x to x-1 on every group, which isn't a group homomorphism in general.
I don't know quite how to prove that every autoequivalence of the category of groups is naturally isomorphic to the identity. I would start with some easier examples, like the category of sets or the category of pointed sets, and then maybe try to use facts about cogroup objects in the category of groups to reduce the problem for groups to the case of pointed sets.
Suppose $F:\mathrm{Grp}\to\mathrm{Grp}$ is an equivalence. The object $\mathbb{Z}\in\mathrm{Grp}$ is a minimal generator (it is a generator, and no proper quotient is also a generator), and this property must be preserved by equivalences. Since there is a unique minimal generator, we can fix an isomorphism $\phi:\mathbb Z\to F(\mathbb Z)$. Now $F$ must preserve arbitrary coproducts, so for all cardinals $\kappa$, the isomorphism $\phi$ induces an isomorphism $\phi_\kappa:L_\kappa\to F(L_\kappa)$, where $L_\kappa$ is the free product of $\kappa$ copies of $\mathbb Z$. In particular, if $1$ is the trivial group, $\phi_0:1\to F(1)$ is an isomorphism.
Next pick a group $G\in\mathrm{Grp}$, and consider a free presentation $L_1\to L_0\to G\to1$, that is, an exact sequence with the $L\_i$ free. (For simplicity, we can take $L\_0=L(G)$ the free group on the underlying set of $G$, and $L_1$ to be the free group on the underlying subset the kernel of the obvious map $L_1\to G$; this eliminates choices) Since $F$ is an equivalence, we have another exact sequence $F(L_1)\to F(L_0)\to F(G)\to F(1)$. Fixing bases for $L_1$ and $L_0$ we can use $\phi$ to construct isomorphisms $L_i\to F(L_i)$ for both $i\in\{0,1\}$. Assuming we can prove the square commutes, one gets an isomorphism $\phi_G:G\to F(G)$—this should not be hard, I guess.
The usual arguments prove then in that case the assignment $G \mapsto \phi_G$ is a natural isomorphism between the identity functor of $\mathrm{Grp}$ and $F$.