My question is motivated by [Are the inner automorphisms the only ones that extend to every overgroup?](https://mathoverflow.net/questions/7782/are-the-inner-automorphisms-the-only-ones-that-extend-to-every-overgroup)


What are the auto-equivalences of the category of groups? What kind of structure do they form?


There are certainly some not-so-trivial examples (which are still trivial if one just focus on a single group), e.g. the functor send an abstract group G to the group G^op with the same underlying set with multiplication g,h ---> hg, where hg is the product of h and g in G. [The notation G^op will make sense if you think G as the category BG]



What are the auto-equivalences of the category of groups?
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 $F$.


