Here's my take on the category with one object whose morphisms are functors \$$\mathbf{M}\to\mathbf{M}\$$: Like Jonathan said, such a functor is entirely determined by where it sends morphisms corresponding to prime numbers (although I would include \$$0\$$ in that list), and we're completely free to decide where to send them (except, because \$$0\cdot n = 0\$$ for all \$$n\$$, we must either send \$$0\$$ to \$$0\$$ or send *everything* to \$$1\$$).

Therefore the set of morphisms in this category can be indexed by \$$\mathbb{N}^\mathbb{N}\cup \\{\ast\\}\$$, infinite sequences of natural numbers together with a special extra element, where a sequence \$$(n_0,n_1,n_2,\dots)\in\mathbb{N}^\mathbb{N}\$$ corresponds to the functor \$$\mathbf{M}\to\mathbf{M}\$$ sending each \$$k\$$th prime \$$p_k\$$ to \$$n_k\$$, and \$$\ast\$$ corresponds to the "send everything to the identity" morphism. All that's left is to describe how these compose.

First, \$$\ast\$$ composed with anything, in either order, is itself. So we just need to worry about how to compose infinite sequences. For that, say we have two sequences \$$(m_0,m_1,m_2,\dots), (n_0,n_1,n_2,\dots)\$$; what is their composite \$$(m_0,m_1,m_2,\dots)\circ(n_0,n_1,n_2,\dots)\$$? Well, let's see where the \$$k\$$th prime \$$p_k\$$ goes. Under the first map it gets sent to \$$n_k\$$. If \$$n_k=0\$$, then the second map will leave it as zero, so under the composite, \$$p_k\mapsto 0\$$. Otherwise, write \$$n_k\$$ as a product of primes \$$p_0^{a_0}p_1^{a_1}p_2^{a_2}\ldots\$$; then \$$p_k\mapsto n_k \mapsto m_0^{a_0}m_1^{a_1}m_2^{a_2}\ldots\$$.

In symbolic form:
$(m_k)_{k\in\mathbb{N}} \circ (n_k)_{k\in\mathbb{N}} = \left(\begin{cases}0&\text{ if }n_k=0\\ m_0^{a_0}m_1^{a_1}m_2^{a_2}\ldots&\text{ if }n_k = p_0^{a_0}p_1^{a_1}p_2^{a_2}\ldots\end{cases}\right)_{k\in\mathbb{N}}.$

It occurs to me that we could also describe how the functors \$$F:\mathbf{M}\to\mathbf{M}\$$ act on morphisms as the *multiplicative functions* from \$$\mathbb{N}\$$ to itself, namely, those functions \$$f:\mathbb{N}\to\mathbb{N}\$$ such that \$$f(1)=1\$$ and \$$f(mn)=f(m)f(n)\$$ for all \$$m,n\in\mathbb{N}\$$. Each non-constant multiplicative function corresponds to an infinite sequence as I described above.

Just please don't call this category \$$\mathbf{P}\$$ and ask me to find its endofunctor monoid.