I was thinking about this question a couple of days ago, and got reminded again by a recent question on graph homomorphisms.


Given a graph $G$, we call two vertices $u,v$ indistinguishable if the map which interchanges $u$ and $v$ and is the identity on the rest of the vertices is an isomorphism. Let $\mathbb{Graph}$ be the underlying quiver of the category of graphs with graph homomorphisms. I.e. the vertices are isomorphism classes of finite graphs, and there are $|Hom(G,H)|$ arrows from $G$ to $H$.


Is it true that $\mathbb{Graph}$ has no pairs of indistinguishable vetices?



Indistinguishable objects in the category of graphs
It's an old result of Lovasz that if $|\mathrm{Hom}(G,X)|=|\mathrm{Hom}(H,X)|$ for all graphs $X$, then $G$ and $H$ are isomorphic. If I understand your quiver correctly, the answer to your question is yes, for the trivial reason that the quiver does not contain a pair of equivalent vertices.


