John, Matthew: I think there is a mild problem with your construction of a non-trivial functor \\(F\colon \mathbf{Set}\to \mathbb{N}\\). You have the map \\(f\colon \emptyset\to \mathbb{R}\\), with \\(F(f)=n\\), and with all other functions going to \\(\mathrm{id}\\). Let \\(t \colon \mathbb{R}\to \\{\ast\\}\\) be the function that send every number to \\(\ast\\). Of course with that domain and that codomain you have no choice. Then we have the composite \\(t\circ f\colon \emptyset \to \(\ast\)\\). We should then have \\(n=\mathrm{id}\circ n = F(t)\circ F(f)= F(t\circ f) =\mathrm{id} \\).

We need to send *all* initial functions to \\(n\\). I guess we have a functor from \\(\mathbf{Set}\\) to the category with two objects \\(a\\) and \\(b\\) and with precisely one non-identity morphism \\(a\to b\\): the empty set gets sent to \\(a\\), everything else goes to \\(b\\). The functor you want from \\(\mathbf{Set}\\) to \\(\mathbb{N}\\) factors through this.

We need to send *all* initial functions to \\(n\\). I guess we have a functor from \\(\mathbf{Set}\\) to the category with two objects \\(a\\) and \\(b\\) and with precisely one non-identity morphism \\(a\to b\\): the empty set gets sent to \\(a\\), everything else goes to \\(b\\). The functor you want from \\(\mathbf{Set}\\) to \\(\mathbb{N}\\) factors through this.