TITLE: Equivalence of Categories and of Their Functor Categories
QUESTION [8 upvotes]: Suppose $A, B, C$ are categories. If $A$ and $B$ are equivalent, is it the case that $C^A$ and $C^B$ are equivalent? Also, is it the case that $A^C$ and $B^C$ are equivalent.
I first conjectured that $C^A$ and $C^B$ are equivalent, but then as I tried to prove the conjecture, I started to realize that it might not be the case. So now I'm trying to find a counterexample. So far I have not been able to find one. Maybe I'm stuck with "too complicated" ideas. I strongly believe that there are very simple counterexamples. Please help. (For more detail, I got to the point where I have a cube that I want to be commutative, but I can only show that 4 of its sides are commutative squares. That's when I realize the other two sides may not necessarily commute.)
And of course, there is the second part to the problem that's not just the dual problem. I would like to know about the special case in which $A$ is a preorder and $B$ is a partial order which is the skeleton of $A$. This may have nothing to do with the validity of the statement in question at all, but it is the setting that originated this question.

REPLY [8 votes]: Recall that $\mathsf{Cat}$ (the category of all small categories) is a $\mathsf{Cat}$-enriched category. Thus we have internal hom functors. Let me spell this out explicitly:
You should be able to check that $A \mapsto [A,C]$ (this is what you write as $C^A$) is a contravariant $\mathsf{Cat}$-endofunctor of $\mathsf{Cat}$ , i.e. that we have functors $[B,A] \to [[A,C],[B,C]]$, $F \mapsto F^*$ which commute with the identity and composition. As such, it preserves equivalences by formal nonsense.
Similarly, $A \mapsto [C,A]$ provides a covariant $\mathsf{Cat}$-endofunctor of $\mathsf{Cat}$, i.e. have functors $[A,B] \to [[C,A],[C,B]]$, $F \mapsto F_*$ which commute with the identity and composition. Again it follows that equivalences are preserved.