Haha. Yes, Owen, I think you've nailed the summary.

> What it sounds like to me is you're thinking about which functors from \$$\mathbf{M}\$$ to itself are *isomorphisms*

Hmm, not quite. I'm not particularly interested in invertible functors, but rather how we can compare two different functors. Consider the example I gave earlier:
> One interesting thing to note is that two functors give isomorphic images if their action on the primes is different only by a permutation of the primes. For instance, if \$$F\$$ sends \$$2\$$ to \$$42\$$ and \$$3\$$ to \$$101\$$, \$$F'\$$ sends \$$3\$$ to \$$42\$$ and \$$2\$$ to \$$101\$$, and both fix all other primes, then they only differ by the permutation \$$(42\ 101)\$$.

Here we have two distinct functors that are not inverses (indeed, are not invertible, since both \$$101\$$ is mapped onto by two elements), but the image categories they produce seem to be isomorphic. I'm hoping that these functors are related by natural transformations (induced by bijections between the primes, yes), but every time I try to understand natural transformations, I come away feeling like I know less than I started with.