I asked [a closely related question](https://math.stackexchange.com/q/3302519/660) on Mathematics Stackexchange but got no answer.

Let $\mathbf1$ be a category with exactly one object and one morphism, and, for any category $\mathcal C$, write $\mathcal C^{\mathcal C}$ for its category of endofunctors.

I [previously](https://mathoverflow.net/q/334266/461) asked if the equivalence $\mathcal C^{\mathcal C}\sim\mathcal C$ implies that $\mathcal C\sim\mathbf1$, but got no complete answer. 

In the meantime I realized that Theorem 3 in *Complete lattices and the generalized Cantor theorem* by Roy O. Davies, Allan Hayes and George Rousseau, published in *Proc. Amer. Math. Soc.* 27 (1971), 253–258, [link](http://www.ams.org/journals/proc/1971-027-02/S0002-9939-1971-0268091-0/S0002-9939-1971-0268091-0.pdf), shows that the implication $\mathcal C^{\mathcal C}\sim\mathcal C\implies\mathcal C\sim\mathbf1$ does hold for posets. 

But the theorem in question is much stronger! Indeed it shows that, for a poset $X$, the existence of an injective morphism $\text{End}(X)\to X$ implies that $X$ is a singleton, and similarly for a surjective morphism $X\to\text{End}(X)$. 

This suggests the following questions:

Let $\mathcal C$ be a category.

> **Question 1** Does the existence of a fully faithful functor $\mathcal C^{\mathcal C}\to\mathcal C$ imply $\mathcal C\simeq\mathbf1$?

> **Question 2** Does the existence of an essentially surjective functor $\mathcal C\to\mathcal C^{\mathcal C}$ imply $\mathcal C\simeq\mathbf1$?

Of course a positive answer to any of these questions would also solve the previously asked [question](https://mathoverflow.net/q/334266/461).