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.

connected. For if $C$ has $\kappa$ many connected components, then $C^C$ has at least $\kappa^\kappa$ many connected components, and the only $\kappa$ such that $\kappa = \kappa^\kappa$ is $\kappa = 1$. Since the identity is connected to the constant functors, there is even a bound on the length of zigzag needed to get from one object to another. $\endgroup$ – Tim Campion Jun 20 at 20:55