> > Each functor corresponds to a morphism and vice versa.
>
> But you've stated that the objects of \$$\mathbf{Mult}\$$ are the morphisms of \$$\mathbf{N}\$$. Is this not a something of a level slip?

I don't think so.

Below I am going to use \$$\bullet\$$ for functorial composition.

For a category \$$\mathcal{C}\$$ an endofunctor \$$F: \mathbf{Mor}(\mathcal{C}) \to \mathbf{Mor}(\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:

$((A \bullet B) \bullet C)(f) = (A \bullet (B \bullet C))(f)$

We also have the *identity functor* \$$\mathbf{1}_{\bullet}\$$ that leaves morphisms alone, so it obeys:

$(\mathbf{1}\_{\bullet} \bullet F)(f) = (F \bullet \mathbf{1}\_{\bullet})(f) = F(f)$

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).

> I would have expected \$$\textbf{Mult}\$$ to be the single-object category with \$$\mathbf{N}\$$ as its object, and functors \$$\mathbf{N} \to \mathbf{N}\$$ as morphisms. After all, the set of endofunctors on \$$\mathbf{N}\$$ forms a commutative monoid by their identification with integer scaling, and monoids are the same as single-object categories.

Well, truth be told I am not sure what the convention is.

I think you could define \$$\textbf{Mult}\$$ either way I think.

Hopefully someone can come along and say definitively what the objects are in the category of endofunctors of a category \$$\mathcal{C}\$$.

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