> **Puzzle 121.** Let \\(\mathbf{N}\\) be the free category on this graph:

> It has one morphism for each natural number. What are all the functors \\(F: \mathbf{N} \to \mathbf{N}\\)?

As Jonathan Castello writes \\(\langle \mathbf{N}, \circ, id_{\star} \rangle \\) isomorphic to the monoid \\(\langle \mathbb{N}, +, 0\rangle\\).

An endofunctor \\(F\\) obeys the law:

\[ F(x \circ y) = F(x) \circ F(y) \]

Hence \\(F\\) is a bit like a [linear map](https://en.wikipedia.org/wiki/Linear_map). We also have

\[ F(id_{\star}) = id_{\star} \]

So we know that \\(F\\) is kind of [*scaling*](https://en.wikipedia.org/wiki/Scaling_(geometry)).

Hence the functors \\(F\\) behave like multiplying constants on \\(\mathbb{N}\\). There is one for each morphism in \\(\mathbb{N}\\).

> What happens when we compose them?

For one, we get a category, which I am going to call \\(\mathbf{Mor}\\)

- The objects are the morphisms of \\(\mathbf{N}\\)
- The identity is the identity functor \\(\mathbf{1}_{\bullet}\\), corresponding to the scaling \\(1 \times \cdot\\)
- Morphisms in this category are functors \\(F : \mathbf{N} \to \mathbf{N}\\)
- Morphism composition is functor composition \\(\bullet\\)

As Jonathan alluded to in [#1](https://forum.azimuthproject.org/discussion/comment/19163/#Comment_19163), this category is like the covariant [Cayley representation](https://en.wikipedia.org/wiki/Cayley%27s_theorem#Proof_of_the_theorem) of \\(\langle \mathbb{N}, \times, 1\rangle\\).

Now that we have \\(\mathbf{Mult}\\), I have another idea...

**Puzzle MD 1**: Consider the functors \\(E : \mathbf{N} \to \mathbf{Mult}\\). What are these functors like?