It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors $\mathsf{T}: \mathsf{Groups} \to \mathsf{Groups}$ with object function $\mathsf{T}(\mathsf{G}) = \mathsf{G}$ for every group $\mathsf{G}$. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?


If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are **not** two such functors?



Two functors from Grp to Grp?
This is an "[evil](http://ncatlab.org/nlab/show/evil)" question, which deserves an evil answer. Pick your favorite pair of an object G0 of Groups and a nontrivial automorphism φ of G0. Define the functor T : Groups → Groups by T(G) = G, T(f) = [φ ∘] f [∘ φ-1] where we compose with φ if the target of f is G0 and compose with φ-1 if the domain of f is G0. This T is clearly different from the identity functor (it acts by φ on Hom(Z, G0) = the underlying set of G0).


This answer isn't very satisfying though because T is naturally isomorphic to the identity. I don't know whether one can find an example where T is not naturally isomorphic to the identity.


