I don't follow. How do you know that for every finite space $Y$, if $\hom(Y, Y) \cong \hom(\mathbb{S}_n, \mathbb{S}_n)$, then $Y \cong \mathbb{S}_n$?

My offhand guess would be 'no', but have you settled this for finite spaces?

Dear Amnon: the downvotes might have been confusing, but I suspect they are connected with an unwritten rule at MO to avoid the appearance of author self-promotion, particularly with regard to books for sale.

@DanPetersen +1. Yes, that is precisely the point (and essentially the point made by Dylan, of course).

Actually, I don't find "non-abelian category" unclear. I can't think of a realistic situation where I would ever use that phrase, except maybe in response to a beginner question, but a category either is abelian or it isn't, and there's only one way (up to canonical iso's, etc.) that a category can be an abelian category. It's not in the same league as equipping a set with a group structure.

On further reflection: what Bugs might mean is that, for example, there is more than one way to endow an object $Z$ with the structure of being a product of objects $X, Y$ (using different choices of projection maps), so that Person A's notion of how $Z$ is a product doesn't match Person B's notion. But that doesn't matter: we say that $F$ preserves products if the canonical map $F(X \times Y) \to FX \times FY$ is an isomorphism. With this understanding of what preservation of (property-like) categorical structure means, all equivalences preserve zero objects, biproducts, and so on.

"Being abelian" is property-like structure (see ncatlab.org/nlab/show/…) that is preserved by any equivalence. One could go through this step by step. Is having a zero object preserved by equivalence? Check. Is having biproducts preserved by equivalence? Check. And so on. Now it's quite true that such property-like structure is not preserved by arbitrary functors. But functors that are equivalences? Certainly.

It's your call. I'm not suggesting you should. Maybe someone would like to say something even more enlightening than the hints given so far in comments; dunno.

Certainly it's the profinite topology, i.e., the topology inherited from the profinite completion $\hat{\mathbb{Z}}$ that appears in the adeles, as pointed out by Chandan Singh Dalawat here: mathoverflow.net/q/42589/2926

Meanwhile, it has an accepted answer over there.

As for which categories of algebras are such that the free functor $Set \to Alg$ is comonadic: these are quite general. See this section in the nLab: ncatlab.org/nlab/show/…