Write $X$ for the set of finitely-supported functions $\mathbb{N} \leftarrow \mathbb{N}$. Then $(X,+,0)$ is the commutative monoid freely generated by $\mathbb{N}$-many generators. So the endomorphisms of $(X,+,0)$ are easy to describe.
Question. What are the endomorphisms of the (commutative, idempotent) monoid $(X \sqcup \{\infty\},\wedge,\infty)?$
Define $\wedge$ by taking a pointwise minimum, as in: $$(g \wedge f)(n) = \mathrm{min}\{g(n),f(n)\},$$ and by asserting that $\infty$ is the identity element for $\wedge$.
Discussion. The fundamental theorem of arithmetic is an isomorphism between $(X,+,0)$ and $(\mathbb{N}_{\geq 1},\times,1).$ By adjoining an extra point, this can be used to get an isomorphism $$(X \sqcup \{\infty\},\wedge,\infty) \cong (\mathbb{N},\gcd,0).$$
Now by user236182's answer here, its a good idea to assign to each $a,b \in \mathbb{N}$ the function $f_{a,b} : \mathbb{N} \leftarrow \mathbb{N}$ defined as follows: $$f_{a,b}(m) = m^a - m^b$$ From his or her answer, we see that if $a$ and $b$ are coprime, then $f_{a,b}$ is an endomorphism, as desired. We can also take pointwise $\mathrm{gcd}$'s of endomorphisms to obtain new endomorphisms. I wonder whether we can get every endomorphism in this way.
