If $\mathcal C^{\mathcal C}$ is equivalent to $\mathcal C$, is $\mathcal C$ necessarily equivalent to a category with one object and one morphism? Let $\mathcal C$ be a category which is equivalent to the category $\mathcal C^{\mathcal C}$ of its endofunctors.

Is $\mathcal C$ necessarily equivalent to a category having exactly one object and one morphism?

A particular case of this question was asked here.
 A: Here is a constraint of a different nature from the ones mentioned so far in the comments. I work in ZFC with one universe $V_\kappa =: Set$. Let $CARD$ denote the class of cardinals $<\kappa$. Below all arithmetic is cardinal arithmetic (not ordinal arithmetic).
Proposition: Let $\mathcal C$ be a category with $\mathcal C \simeq \mathcal C^{\mathcal C}$. Assume that


*

*The number of isomorphism classes of objects of $\mathcal C$ is $< 2^\kappa$

*$\mathcal C$ is locally small (its homsets are of size $<\kappa$)

*$\mathcal C$ has all small powers and all small copowers.
Then $\mathcal C$ is a preorder.
Remarks:


*

*There are two important cases where $\mathcal C$ has all small powers and all small copowers:


*

*if $\mathcal C$ has all small products and all small coproducts.

*if $\mathcal C$ is a preorder
So the proposition could be stated as an "if and only if".

*Of course, the existence of the powers and copowers should be viewed as the"substantive" hypothesis, and the size restrictions as "technical".

*I think these hypotheses are general enough to make the case that no "nice large category" satisfies $\mathcal C \simeq \mathcal C^{\mathcal C}$, but of course it would be interesting to weaken the hypothesis. That would probably require quite different methods not reducing to the existence of lots of $Set$-functors as below. There's another paper of Koubek on contravariant $Set$-functors, which might allow one to assume only the existence of powers without assuming the existence of copowers (or dually).
Proof:
For any $F: Set \to Set$, and $A \in \mathcal C$, we may form a functor $\Phi_{F,A}: \mathcal C \to \mathcal C$, $C \mapsto \amalg_{x \in F(\mathcal C(A,C))} A$. Suppose for contradiction that $\mathcal C$ is not a preorder. Fix $A,B$ such that
$$\lambda := |\mathcal C(A,B)| \geq 2$$
Let $\mathcal A$ be the class of (small) cardinals of the form $\alpha = \lambda^\mu$, and let $\mathcal F$ be the class of all weakly increasing functions $f: \mathcal A \to CARD$ such that


*

*$f(\alpha) \geq 2^\alpha$ for all $\alpha \in \mathcal A$. 

*For all $\alpha \in \mathcal A$, if $\lambda^{f(\alpha)} = \lambda^\beta$, then $f(\alpha) \leq \beta$.
Clearly $|\mathcal F| = 2^\kappa$. By a theorem of Koubek [1], for every $f \in \mathcal F$, there is a functor  $F_f: Set \to Set$ with $|F_f(X)| = f(|X|)$ for all $X$ with $|X| \in \mathcal A$. Because $\mathcal C$, and hence $\mathcal C^{\mathcal C}$, has only $< 2^\kappa$ many isomorphism classes of objects, there are $f_1 \neq f_2 \in \mathcal F$ such that $\Phi_{F_1,A} \cong \Phi_{F_2,A}$ (where $F_i = F_{f_i}$).  Since $f_1 \neq f_2$, there is a cardinal $\mu$ such that $f_1(\lambda^\mu) \neq f_2(\lambda^\mu)$. We have that $\Phi_{F_1,A}(\prod_\mu B) \cong\Phi_{F_2,A}(\prod_\mu B)$, i.e. $\amalg_{f_1(\lambda^\mu)} A \cong \amalg_{f_2(\lambda^\mu)} A$. Homming into $B$, we find that $\mathcal C(A,B)^{f_1(\lambda^\mu)} \cong \mathcal C(A,B)^{f_2(\lambda^\mu)}$. But by the second condition in the definition of $\mathcal F$, this implies that $f_1(\lambda^\mu) = f_2(\lambda^\mu)$, contradicting the choice of $\mu$.
[1] Koubek, Václav. "Set functors." Commentationes Mathematicae Universitatis Carolinae 012.1 (1971): 175-195. http://eudml.org/doc/16420, Prop 2.4, bottom of p. 183 (p.10 of the pdf) -- note that bizarrely in this paper that means section 4, second proposition, and is also different from Lemma 2.4.
A: This is a partial answer. I tried to mimic the proof of Theorem 3 in 
[1] 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.
(The notation used in this answer is different from the one in the question. I'll stick as much as possible to the notation and terminology of [1].) 
All categories in this post are small.
Let $A$ and $B$ be categories, and let $b_0,b_1$ be two objects of $B$ such that there is a morphism $b_0\to b_1$ but no morphism $b_1\to b_0$.
We claim
(1) $B^B$ is not equivalent to $B$,
(2) there is no full functor $B^A\to A$ and no essentially surjective functor $A\to B^A$.
It suffices to prove (2).
For any category $C$ write $C_0$ for the set of objects of $C$. 
Let $C$ and $D$ be categories. Say that a map $F:C_0\to D_0$ is a weak monomorphism if the existence of morphisms $c\to c'$ and $F(c')\to F(c)$ implies that of a morphism $c'\to c$.
It is clear that the map $C_0\to D_0$ induced by a full functor is a weak monomorphism. It is also clear that the existence of an essentially surjective functor  $D\to C$ implies that of a weak monomorphism $C_0\to D_0$. 
We claim 
(3) There is no weak monomorphism $(B^A)_0\to A_0$.
It suffices to prove (3).
Let $2$ be the ordinal $\{0,1\}$ viewed as a category. We claim
(4) There is no weak monomorphism $(2^A)_0\to A_0$.
Let us show that (4) implies (3). 
Let $F:(B^A)_0\to A_0$ be a weak monomorphism, let $J:2\to B$ be a functor mapping $i$ to $b_i$ for $i=0,1$ (such a functor exists by assumption), and define $F':(2^A)_0\to A_0$ by $F'(G):=F(J\circ G)$. It is straightforward to check (using the assumption that there is no morphism $b_1\to b_0$) that $F'$ is a weak monomorphism (details left to the reader). This proves that (4) implies (3). 
It suffices to prove (4). 
Say that a subset $R$ of $A_0$ is a right ideal if the conditions $a\in R$ and there is a morphism $a\to a'$ imply $a'\in R$. The right ideals form a complete lattice $\mathcal R$ order isomorphic to $(2^A)_0$. (The order on $\mathcal R$ is given by inclusion.) 
Thus it suffices to show that there is no weak monomorphism $\mathcal R\to A_0$.
Let $\phi:\mathcal R\to A_0$ be a map. Define the map $f:\mathcal R\to \mathcal R$ by letting $f(R)$ be the right ideal generated by $\phi(R)$. By the corollary to Theorem 1 in [1] there is an $R$ in $\mathcal R$ such that 
$$
f(R)\le\bigcup_{S>R}f(S).
$$ 
As we have $\phi(R)\in f(R)$, this implies $\phi(R)\in f(S)$ for some $S$ in $\mathcal R$ with 
(5) $R<S$, 
and thus (by definition of $f(S)$) the existence of a morphism 
(6) $\phi(S)\to\phi(R)$. 
Now (5) and (6) imply that $\phi$ is not a weak monomorphism.
