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(Edited to reflect your edit to the question!)

The answer to your original statement (without the "nonempty" assumption) is no because we can let $A=\varnothing$ and $B=\ast$. Their endofunctor categories are each discrete with one object, but the categories themselves are not equivalent.

The question becomes a lot trickier when you add "nonempty" to the statement, but I think the answer is still no. I believe the following is a counterexample:

Let $G$ be an abelian group such that $G\times G\cong G$ and $\text{End}(G,G)$ is infinite. (So, for example, let $G=\Pi_{n\geq 1} \mathbb{Z}$). Let $C$ be the one object category corresponding to $G$. Notice that $C$ is not equivalent to $C\sqcup C$ because $C$ has just one object and $C\sqcup C$ has two non-isomorphic objects. But let's compare their endofunctor categories.

First, let's describe $C^C$. The objects in $C^C$ are precisely the homomorphisms $G\rightarrow G$. Each morphism in $C^C$, i.e., each natural transformation $\phi\Rightarrow \psi$, is a choice of $g\in G$ such that $g\phi(h)=\psi(h)g$ for all $h$, but since $G$ is abelian, that means $\phi=\psi$. Thus, $C^C\cong \sqcup_{\text{End}(G,G)} C$. Now, observe that $(C\sqcup C)^C\cong C^C \sqcup C^C$ since $C$ has one object. Putting these together and using the fact that $C^{(-)}$ sends colimits to limits, we have

$(C\sqcup C)^{C\sqcup C}\cong (C\sqcup C)^C\times (C\sqcup C)^C \cong (C^C \sqcup C^C)\times (C^C\sqcup C^C)\cong C^C \times C^C$,

where the last isomorphism comes from $C^C$ being an infinite coproduct. But then

$C^C\times C^C \cong (\sqcup_{\text{End}(G,G)} C) \times (\sqcup_{\text{End}(G,G)} C)\cong \sqcup_{\text{End}(G,G)\times \text{End}(G,G)} (C\times C)\cong \sqcup_{\text{End}(G,G)} C\cong C^C$

(Edited to reflect your edit to the question!)

The answer to your original statement (without the "nonempty" assumption) is no because we can let $A=\varnothing$ and $B=\ast$. Their endofunctor categories are each discrete with one object, but the categories themselves are not equivalent.

The question becomes a lot trickier when you add "nonempty" to the statement, but I think the answer is still no. I believe the following is a counterexample:

Let $G$ be an abelian group such that $G\times G\cong G$ and $\text{End}(G)$ is infinite. (So, for example, let $G=\Pi_{n\geq 1} \mathbb{Z}$). Let $C$ be the one object category corresponding to $G$. Notice that $C$ is not equivalent to $C\sqcup C$ because $C$ has just one object and $C\sqcup C$ has two non-isomorphic objects. But let's compare their endofunctor categories.

First, let's describe $C^C$. The objects in $C^C$ are precisely the homomorphisms $G\rightarrow G$. Each morphism in $C^C$, i.e., each natural transformation $\phi\Rightarrow \psi$, is a choice of $g\in G$ such that $g\phi(h)=\psi(h)g$ for all $h$, but since $G$ is abelian, that means $\phi=\psi$. Thus, $C^C\cong \sqcup_{\text{End}(G)} C$. Now, observe that $(C\sqcup C)^C\cong C^C \sqcup C^C$ since $C$ has one object. Putting these together and using the fact that $C^{(-)}$ sends colimits to limits, we have

$(C\sqcup C)^{C\sqcup C}\cong (C\sqcup C)^C\times (C\sqcup C)^C \cong (C^C \sqcup C^C)\times (C^C\sqcup C^C)\cong C^C \times C^C$,

where the last isomorphism comes from $C^C$ being an infinite coproduct. But then

$C^C\times C^C \cong (\sqcup_{\text{End}(G)} C) \times (\sqcup_{\text{End}(G)} C)\cong \sqcup_{\text{End}(G)\times \text{End}(G)} (C\times C)\cong \sqcup_{\text{End}(G)} C\cong C^C$

(Note: In this example, the endofunctor categories are in fact isomorphic, not just equivalent.)

(Edited to reflect your edit to the question!)

The answer to your original statement (without the "nonempty" assumption) is no because we can let $A=\varnothing$ and $B=\ast$. Their endofunctor categories are each discrete with one object, but the categories themselves are not equivalent.

The question becomes a lot trickier when you add "nonempty" to the statement, but I think the answer is still no. I believe the following is a counterexample:

Let $G$ be an abelian group such that $G\times G\cong G$ and $\text{End}(G,G)$ is infinite. (So, for example, let $G=\Pi_{n\geq 1} \mathbb{Z}$). Let $C$ be the one object category corresponding to $G$. Notice that $C$ is not equivalent to $C\sqcup C$ because $C$ has just one object and $C\sqcup C$ has two non-isomorphic objects. But let's compare their endofunctor categories.

First, let's describe $C^C$. The objects in $C^C$ are precisely the homomorphisms $G\rightarrow G$. Each morphism in $C$, i.e., each natural transformation $\phi\Rightarrow \psi$, is a choice of $g\in G$ such that $g\phi(h)=\psi(h)g$ for all $h$, but since $G$ is abelian, that means $\phi=\psi$. Thus, $C^C\cong \sqcup_{\text{End}(G,G)} C$. Now, observe that $(C\sqcup C)^C\cong C^C \sqcup C^C$ since $C$ has one object. Putting these together and using the fact that $C^{(-)}$ sends colimits to limits, we have

$(C\sqcup C)^{C\sqcup C}\cong (C\sqcup C)^C\times (C\sqcup C)^C \cong (C^C \sqcup C^C)\times (C^C\sqcup C^C)\cong C^C \times C^C$,

where the last isomorphism comes from $C^C$ being an infinite coproduct. But then

$C^C\times C^C \cong (\sqcup_{\text{End}(G,G)} C) \times (\sqcup_{\text{End}(G,G)} C)\cong \sqcup_{\text{End}(G,G)\times \text{End}(G,G)} (C\times C)\cong \sqcup_{\text{End}(G,G)} C\cong C^C$

(Edited to reflect your edit to the question!)

The answer to your original statement (without the "nonempty" assumption) is no because we can let $A=\varnothing$ and $B=\ast$. Their endofunctor categories are each discrete with one object, but the categories themselves are not equivalent.

The question becomes a lot trickier when you add "nonempty" to the statement, but I think the answer is still no. I believe the following is a counterexample:

Let $G$ be an abelian group such that $G\times G\cong G$ and $\text{End}(G,G)$ is infinite. (So, for example, let $G=\Pi_{n\geq 1} \mathbb{Z}$). Let $C$ be the one object category corresponding to $G$. Notice that $C$ is not equivalent to $C\sqcup C$ because $C$ has just one object and $C\sqcup C$ has two non-isomorphic objects. But let's compare their endofunctor categories.

First, let's describe $C^C$. The objects in $C^C$ are precisely the homomorphisms $G\rightarrow G$. Each morphism in $C^C$, i.e., each natural transformation $\phi\Rightarrow \psi$, is a choice of $g\in G$ such that $g\phi(h)=\psi(h)g$ for all $h$, but since $G$ is abelian, that means $\phi=\psi$. Thus, $C^C\cong \sqcup_{\text{End}(G,G)} C$. Now, observe that $(C\sqcup C)^C\cong C^C \sqcup C^C$ since $C$ has one object. Putting these together and using the fact that $C^{(-)}$ sends colimits to limits, we have

$(C\sqcup C)^{C\sqcup C}\cong (C\sqcup C)^C\times (C\sqcup C)^C \cong (C^C \sqcup C^C)\times (C^C\sqcup C^C)\cong C^C \times C^C$,

where the last isomorphism comes from $C^C$ being an infinite coproduct. But then

$C^C\times C^C \cong (\sqcup_{\text{End}(G,G)} C) \times (\sqcup_{\text{End}(G,G)} C)\cong \sqcup_{\text{End}(G,G)\times \text{End}(G,G)} (C\times C)\cong \sqcup_{\text{End}(G,G)} C\cong C^C$

The answer to your precise statement is definitely no because we can let $A=\varnothing$ and $B=\ast$. Their endofunctor categories are each discrete with one object, but the categories themselves are not equivalent.

The question becomes a lot trickier when you add "nonempty" to the statement, but I think the answer is still no. I believe the following is a counterexample:

Let $G$ be an abelian group such that $G\times G\cong G$ and $\text{End}(G,G)$ is infinite. (So, for example, let $G=\Pi_{n\geq 1} \mathbb{Z}$). Let $C$ be the one object category corresponding to $G$. Notice that $C$ is not equivalent to $C\sqcup C$ because $C$ has just one object and $C\sqcup C$ has two non-isomorphic objects. But let's compare their endofunctor categories.

First, let's describe $C^C$. The objects in $C^C$ are precisely the homomorphisms $G\rightarrow G$. Each morphism in $C$, i.e., each natural transformation $\phi\Rightarrow \psi$, is a choice of $g\in G$ such that $g\phi(h)=\psi(h)g$ for all $h$, but since $G$ is abelian, that means $\phi=\psi$. Thus, $C^C\cong \sqcup_{\text{End}(G,G)} C$. Now, observe that $(C\sqcup C)^C\cong C^C \sqcup C^C$ since $C$ has one object. Putting these together and using the fact that $C^{(-)}$ sends colimits to limits, we have

$(C\sqcup C)^{C\sqcup C}\cong (C\sqcup C)^C\times (C\sqcup C)^C \cong (C^C \sqcup C^C)\times (C^C\sqcup C^C)\cong C^C \times C^C$,

where the last isomorphism comes from $C^C$ being an infinite coproduct. But then

$C^C\times C^C \cong (\sqcup_{\text{End}(G,G)} C) \times (\sqcup_{\text{End}(G,G)} C)\cong \sqcup_{\text{End}(G,G)\times \text{End}(G,G)} (C\times C)\cong \sqcup_{\text{End}(G,G)} C\cong C^C$

(Edited to reflect your edit to the question!)

The answer to your original statement (without the "nonempty" assumption) is no because we can let $A=\varnothing$ and $B=\ast$. Their endofunctor categories are each discrete with one object, but the categories themselves are not equivalent.

The question becomes a lot trickier when you add "nonempty" to the statement, but I think the answer is still no. I believe the following is a counterexample:

Let $G$ be an abelian group such that $G\times G\cong G$ and $\text{End}(G,G)$ is infinite. (So, for example, let $G=\Pi_{n\geq 1} \mathbb{Z}$). Let $C$ be the one object category corresponding to $G$. Notice that $C$ is not equivalent to $C\sqcup C$ because $C$ has just one object and $C\sqcup C$ has two non-isomorphic objects. But let's compare their endofunctor categories.

First, let's describe $C^C$. The objects in $C^C$ are precisely the homomorphisms $G\rightarrow G$. Each morphism in $C$, i.e., each natural transformation $\phi\Rightarrow \psi$, is a choice of $g\in G$ such that $g\phi(h)=\psi(h)g$ for all $h$, but since $G$ is abelian, that means $\phi=\psi$. Thus, $C^C\cong \sqcup_{\text{End}(G,G)} C$. Now, observe that $(C\sqcup C)^C\cong C^C \sqcup C^C$ since $C$ has one object. Putting these together and using the fact that $C^{(-)}$ sends colimits to limits, we have

$(C\sqcup C)^{C\sqcup C}\cong (C\sqcup C)^C\times (C\sqcup C)^C \cong (C^C \sqcup C^C)\times (C^C\sqcup C^C)\cong C^C \times C^C$,

where the last isomorphism comes from $C^C$ being an infinite coproduct. But then

$C^C\times C^C \cong (\sqcup_{\text{End}(G,G)} C) \times (\sqcup_{\text{End}(G,G)} C)\cong \sqcup_{\text{End}(G,G)\times \text{End}(G,G)} (C\times C)\cong \sqcup_{\text{End}(G,G)} C\cong C^C$