Equivalence of categories of abelian presheaves reflects isomorphisms of rigid categories?

Question

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 $D$?

Answer 1

No, this is false. Let $C$ be the monoid $\lbrace 1,\ldots, 2^n\rbrace$ with $\max$ as the operation and let $D$ be the power set of $\lbrace 1,\ldots, n\rbrace$ with $\cup$ as the operation. These are both join semilattices with identity of cardinality $2^n$. The integral monoid rings of two finite join semilattices with identity of the same cardinality are isomorphic by Mobius inversion. Of course there are no isomorphisms other than identities here. More precisely, they are both isomorphic to a direct product of $2^n$ copies of $\mathbb Z$. Of course for a monoid, abelian presheaves are the same as modules over the integral monoid ring.

See this paper of Solomon for the proof the algebras are isomorphic.

Added. Harry wanted an example where the categories are Cauchy complete. The Cauchy completions of these monoids are not isomorphic and are rigid. It is well known that for monoids $M$ and $N$ one has that $\mathbf{Set}^{M^{op}}\cong \mathbf{Set}^{N^{op}}$ if and only if there is an idempotent $e\in M$ with $MeM=M$ and $eMe\cong N$. In particular, if $M$ is finite, then one must have $e=1$ and so $M\cong N$. The two monoids I gave above are not isomorphic.