I don't 100% follow the git metaphor either... it sounds like Jonathan is saying "Don't edit posts people may be responding to! It's like this thing you can do in git which is bad!" And Matthew is saying "Sorry it was confusing, and by the way sometimes that bad git thing is the best option." So I think everybody is good now?

Anyway, Jonathan wrote:
> I think you have a better handle on this problem than I do... Do you have any thoughts on the possibility I mentioned about having equivalent functors? It seems like we essentially have a "change of base" property taking us between certain functors, mediated by bijections on the set of prime numbers.

What it sounds like to me is you're thinking about which functors from \\(\mathbf{M}\\) to itself are *isomorphisms*: I think that the answer is exactly those functors that just permute the primes, corresponding to sequences in \\(\mathbb{N}^\mathbb{N}\\) consisting of all the primes in some order. Then you could regard two functors \\(\mathbf{M}\to\mathbf{M}\\) as equivalent if you can get from one to the other by precomposing with an isomorphism, and I *think* that's the notion of equivalence you described above.

In general, if you have two objects \\(A\\) and \\(B\\) in a category, the set of morphisms \\(\mathrm{Hom}(A,B)\\) between them gets two group actions automatically: the group of automorphisms of \\(A\\) (isomorphisms from \\(A\\) to itself) acts on the set by precomposition, and the group of automorphisms of \\(B\\) acts by postcomposition. Sometimes it does make sense to look at the orbits of one or both group actions, and then we're regarding morphisms as equivalent in the way you describe.

Soon, I think, John will tell us about another kind of equivalence between functors: *natural isomorphisms*. Instead of being related by pre- or postcomposition by an invertible functor, two functors are naturally isomorphic if they're related by something that looks more like conjugation.