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.

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.