(**EDIT:** You deleted Puzzle 122! But I liked that one! :'( )

**Puzzle 121:** The free category on this graph is a monoid isomorphic to \\(\langle \mathbb{N}, + \rangle\\), so endofunctors will be monoid endomorphisms. That means functors \\(\mathbf{N} \to \mathbf{N}\\) identify submonoids of \\(\langle \mathbb{N}, + \rangle\\), i.e. the homomorphic images of these functors. It is a theorem that every monoid homomorphism over a finitely presented monoid is determined by where the generators are sent, so it suffices to consider where \\(1\\) is mapped. As it happens, we get a monoid homomorphism for every natural number \\(n\\), determined by \\(1 \mapsto n\\) (i.e., \\(x \mapsto n x\\)). These are our desired functors. (It's worth noting that when \\(n = 1\\), we get the identity functor.)

Composing these functors ends up multiplying their associated scaling factors. In other words, \\(\langle \mathbb{N}, \cdot \rangle\\) arises as the class of functors on \\(\langle \mathbb{N}, + \rangle\\).

**Puzzle 122\\({}^\ast\\):** \\(F\\) will pick a set and a class of automorphisms on that set that is closed under composition, again giving a monoid. Since \\(\mathbf{N}\\) is finitely generated, \\(F\\) is fully determined by where it sends \\(s\\). If \\(F(z)\\) is a finite set, then \\(F(s)\\) must have finite order, i.e. \\((F(s))^{(k)} = \mathrm{id}\\) for some \\(k\\). This means that \\(F(\mathbf{N})\\) is isomorphic to the finite group \\(\mathbb{Z}_k\\) for some \\(k\\), if \\(F(z)\\) is finite. If \\(F(z)\\) is infinite, then we gain the additional opportunity for \\(F(s)\\) to have infinite order. We can do this by mapping \\(z \to \mathbb{Z}\\) and \\(s \to \lambda x. x + 1\\); then we recover a structure isomorphic to \\(\langle \mathbb{N}, + \rangle\\).

Every image of a functor \\(F: \mathbf{N} \to \mathbf{Set}\\) is isomorphic to one of the cases listed above; so while there may be uncountably many specific functors of this kind, there is only a countably infinite set of classes that produce different structures.

Incidentally, the finite set cases above give the same result as if we had imposed an equation on our finite presentation, namely that \\(s^k = \mathrm{id}\\) for some \\(k\\). Is it a general rule that for every finite presentation of a category with equations, there is a suitable functor from the free category that imposes these equations?