Recall that we call a category *rigid* if it contains no non-identity isomorphisms. Let $\mathbf{rig}$ denote the full 2-subcategory of $\mathbf{Cat}$ spanned by the small rigid categories. It is easy to see that a functor in $\mathbf{rig}$ is an equivalence of categories if and only if it is an isomorphism.

Then consider the 2-functor $\mathbf{Ab}(\cdot)=\operatorname{Hom}_{\mathbf{Cat}}((\cdot)^{\operatorname{op}},\mathbf{Ab}):\mathbf{rig}^{\operatorname{op}}\to \mathbf{Cat}$ sending a small rigid category to its associated category of abelian presheaves. Then if there is an equivalence of categories $\mathbf{Ab}(C) \simeq \mathbf{Ab}(D)$, where $C$ and $D$ are rigid, does this imply that $C$ is isomorphic to $D$?