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My question is motivated by 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]

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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 $F$.

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    $\begingroup$ I think there are some subtleties in this approach. Let's say we've shown that F(Z) is abstractly isomorphic to Z. Since Z has a nontrivial automorphism, there are two isomorphisms ϕ we might choose. If we pick the wrong one, we won't be able to extend it to a natural isomorphism between the identity and F, for the reason I described in my answer. So there must be more work involved in constructing ϕ_G. $\endgroup$
    – Reid Barton
    Commented Dec 5, 2009 at 6:00
  • $\begingroup$ @Reid: automorphisms of the identity functor and autoequivalences of the category are different things, and your argument applies to the first. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 5, 2009 at 20:27
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    $\begingroup$ What I am saying is that you are trying to construct an isomorphism between F and the identity, by extending an isomorphism between F(Z) and Z--but there are two isomorphisms of the latter type, and only one of the former, so you might go wrong! $\endgroup$
    – Reid Barton
    Commented Jan 6, 2010 at 21:05
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    $\begingroup$ (And there is only one isomorphism of the former type precisely because the identity functor has no automorphisms.) $\endgroup$
    – Reid Barton
    Commented Jan 6, 2010 at 21:08
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    $\begingroup$ No; any isomorphism phi : Z -> F(Z) will induce isomorphisms phi_k : L_k -> F(L_k). For instance if F is the identity and phi is the multiplication by -1 map, then phi_k will be the map sending each generator of the free group to its inverse. This is natural with respect to homomorphisms between free groups induced from maps of sets of their generators, but not with respect to arbitrary group homomorphisms. So the assumption "we can prove the square commutes" is actually false--or rather, it requires a more complex argument that might go back and change phi. $\endgroup$
    – Reid Barton
    Commented Jan 12, 2010 at 16:29
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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".

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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.

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    $\begingroup$ This is on page 31 of Peter Freyd's "Abelian Categories" book. $\endgroup$
    – Gonçalo Marques
    Commented Dec 4, 2009 at 19:44
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    $\begingroup$ You should make that an answer to the question! $\endgroup$
    – Reid Barton
    Commented Dec 5, 2009 at 3:19

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