Monoids where $\operatorname{Hom}(M,M) \cong M$ What are some examples of monoids where $\operatorname{End}(M) \cong M$? Is there a nice characterization of such monoids? E.g., they will necessarily have a zero element, since $\forall x (x \mapsto 1_M)$ is a zero of $M^M$ (so no groups satisfy $G^G \cong G$).
Examples:


*

*$\{1\}$

*$(\mathbb{F}_2, \cdot)$


Non-examples: 


*

*$(\mathbb N, +) \mapsto (\mathbb N, \cdot)$

*$(\mathbb F_2, +_2) \mapsto (\mathbb F_2, \cdot)$


Partial answer. If $M$ is a finite monoid such that $\operatorname{End}(M) \cong M$, then either $M = \{1\}$ or $M = \{0, 1\}$ with the usual multiplication of integers.
Proof. Let $G$ be the group of invertible elements of $M$. Since $M$ is finite, $M - G$ is an ideal of $M$. Let $E(M)$ be the set of idempotents of $M$. For each $e \in E(M)$, let $\bar e$ be the endomorphism of $M$ defined by
$$
\bar e(x) = \begin{cases} 
1 & \text{if $x \in G$}\\
e & \text{otherwise}
\end{cases}
$$
Then $\bar e \in E(\operatorname{End}(M))$ and moreover $\bar e \bar f = \bar e$ for all $e, f \in E(M)$. Thus the map $e \to \bar e$ is an injection from $E(M)$ to $E(\operatorname{End}(M))$. Since $M$ and $\operatorname{End}(M)$ are isomorphic, this injection is a bijection. In particular, there exists an idempotent $e$ in $M$ such that $\bar e$ is the identity map on $M$. Since the range of $\bar e$ is $\{1, e\}$, this means that $M = \{1, e\}$, which gives the two possible solutions.
