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 if $G$ and $H$ are indistinguishable vertices of $\mathbb{Graph}$, then $G$ and $H$ are equivalent themselves? Here I'm calling two graphs equivalent if they can be transformed one into another by a sequence of identifications of indistinguishable vertices.

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?

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 if $G$ and $H$ are indistinguishable verties of $\mathbb{Graph}$ then $G$ and $H$ are equivalent themselves. Where I'm calling two graphs equivalent if they can be transformed into one another by a sequence of identification of indistinguishable vertices.

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 if $G$ and $H$ are indistinguishable vertices of $\mathbb{Graph}$, then $G$ and $H$ are equivalent themselves? Here I'm calling two graphs equivalent if they can be transformed one into another by a sequence of identifications of indistinguishable vertices.