> For a category \$$\mathcal{C}\$$ an endofunctor \$$F: \mathbf{Hom}(\mathcal{C}) \to \mathbf{Hom}(\mathcal{C})\$$ maps morphisms to morphisms.
>
> If I have three endofunctors \$$A,B,C\$$ and a morphism \$$f\$$ then we have associativity of functorial composition:

So we have \$$A, B, C : \mathbf{Hom}(\mathcal{C}) \to \mathbf{Hom}(\mathcal{C})\$$. Are there endofunctors with any other signature?

I think the difference between our constructions amounts to the difference between [Puzzle 99 and Puzzle 100](https://forum.azimuthproject.org/discussion/2198/lecture-34-chapter-3-categories/p1). In Puzzle 100, we can _label_ arrows from one object to another with particular endomorphisms (group elements), but _every_ object has some out-arrow labeled with this element. There isn't a one-to-one correspondence between endomorphisms and arrows, because we're applying \$$F\$$ specially to each object and relating the single input to the single output.

Put differently, if we take each \$$s^{k}\$$ as an object in our new category, and we induce all arrows \$$s^k \to s^{nk}\$$ for each endofunctor \$$n\$$, we end up with a partial order (except for zero). Indeed, this is the divisibility lattice on the positive natural numbers, augmented with a lonely \$$0\$$ object that behaves like the single-object category I described, with one endomorphism for each \$$n\$$.

(**EDIT:** Some confusion might arise from the fact that we often construct the naturals by \$$z, s(z), s(s(z)), \cdots\$$, and we have a morphism \$$s\$$ and an object \$$z\$$ in evidence here. But \$$s(z)\$$ is a nonsensical construction in \$$\textbf{N}\$$! Morphisms are not applied to objects. Objects represent the domain and codomain of morphisms. If \$$z\$$ were a set, then we could take \$$x \in z\$$ and have \$$s(x) \in z\$$.)

> I don't think this is my construction. Saunder's MacLane defines exactly the same thing on page 128 of [*Categories for the Working Mathematician* (1968)](https://books.google.com/books?id=MXboNPdTv7QC&pg=PA138&lpg=PA138&dq=%22monoid+in+the+category+of+endofunctors%22+mac+lane&source=bl&ots=feQWTkH2Uw&sig=tv-1JwaMOygKGmFE2vM2FhJVS9o&hl=en&ei=5iWsTJCkBIPSsAPQwJ36Aw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q&f=false).

He seems to be doing something with a single fixed endofunctor \$$T\$$ and some natural transformations \$$\eta, \mu\$$, the latter of which I have no familiarity with. We have a whole family of endofunctors and no obvious maps between any of them.

> I don't think it makes a big difference for the puzzle I invented...

If \$$\textbf{Mult}\$$ is the divisibility lattice augmented with zero, then we have essentially two classes of functor \$$F : \mathbf{N} \to \mathbf{Mult}\$$. One functor maps \$$z\$$ onto a non-zero element, forcing all morphisms onto the identity morphism. It doesn't much matter which number we map onto; the image of the functor will be a single object with its identity morphism. The other functors are covered by the case below.

If \$$\textbf{Mult}\$$ is the single-object category with functors \$$\mathbf{N} \to \mathbf{N}\$$ as morphisms, then functors \$$\mathbf{N} \to \mathbf{Mult}\$$ are monoid homomorphisms between \$$\langle \mathbb{N}, + \rangle\$$ and \$$\langle \mathbb{N}, \cdot \rangle\$$. It suffices to consider where \$$1\$$ is mapped to. If \$$1\$$ maps to \$$n\$$ for a choice of \$$n\$$, then \$$k\$$ maps to \$$n^k\$$. In other words, we get exponentiation.

If we do this again, what do we get? There's a [name](https://en.wikipedia.org/wiki/Hyperoperation#Examples) for what's happening here.