Given a monoid $M$ we define the endomorphism monoid $\mathrm{End}(M)$ of $M$ to be the monoid of homomorphisms $M\to M$ along with composition as the operation. My query is whether or not the endomorphism monoid preserves products in the sense that: $$\mathrm{End}(M_1\times M_2)\cong\mathrm{End}(M_1)\times\mathrm{End}(M_2)$$ for monoids $M_1$ and $M_2$. If so, why and if not, why does it fail and are there some restrictions which we may put on $M_1$ and $M_2$ which make the statement above true?
Thank you!
EDIT:
This question was taken down, so I will edit it. My motivation for this is purely for fun: trying to classify monoids such that $$\mathrm{End}(M)\cong M$$ and I was hoping if that if I could get some background information and facts about endomorphism monoids, I could get closer to determining how to classify such monoids.
Apologies for the poorly posed old question: I'm still new here!
Show that $\operatorname{End}(M_1\times M_2)$ contains a sub-monoid $\mathcal N,$ isomorphic to $\operatorname{End}(M_1)\times \operatorname{End}(M_2).$
But there can be more elements. For example, if $M_2=M_1=M,$ there is endomorphism on $M\times M$, $(m,n)\to(n,m)$ which is different from the ones in the $\mathcal N.$
In particular, if $M$ is finite and $|M|>1,$ then $\operatorname{End}(M\times M)$ is finite and larger than $\operatorname{End}(M)\times\operatorname{End}(M).$
A simple example is $M=(\mathbb Z/2\mathbb Z,+).$ Then $\operatorname{End}(M)$ has two elements, but $\operatorname{End}(M\times M)$ has sixteen elements.