(Crossposted from [math.SE](https://math.stackexchange.com/questions/99784/morita-equivalence-of-acyclic-categories).)


Call a category *acyclic* if only the identity morphisms are invertible and the endomorphism monoid of every object is trivial. Let $C, D$ be two finite acyclic categories. Suppose that they are *Morita equivalent* in the sense that the abelian categories $\text{Fun}(C, \text{Vect})$ and $\text{Fun}(D, \text{Vect})$ are equivalent (where $\text{Vect}$ is the category of vector spaces over a field $k$, say algebraically closed of characteristic zero). Does it then follow that $C, D$ are equivalent? (If so, can we drop the finiteness condition?) 


Without the acyclic condition this is false; for example, if $G$ is a finite group regarded as a one-object category, $\text{Fun}(G, \text{Vect})$ is completely determined by the number of conjugacy classes of $G$, and it is easy to write down pairs of nonisomorphic finite groups with the same number of conjugacy classes (take, for example, any nonabelian group $G$ with $n < |G|$ conjugacy classes and $\mathbb{Z}/n\mathbb{Z}$). 


On the other hand, Mariano's answer on math.SE shows that the result is true if $C, D$ are either both taken to be free categories on graphs or both taken to be posets by basic results in the representation theory of quivers. (This and idle curiosity are the main motivation for the question.) 



Morita equivalence of acyclic categories
The answer to your question is "no". A counterexample is given in this [paper](https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-26/issue-1/The-isomorphism-problem-for-incidence-algebras-of-M%C3%B6bius-categories/10.1215/ijm/1256046901.full) by Leroux in Example 1.6.


