Are the distributions of uniform spanning trees unique up to graph isomorphism?

Question

I've tried to find any publications on this unsuccessfully, but maybe someone here has some ideas where to look.

Consider a directed graph $G = (V_G, A_G)$ of $N$ nodes and $M$ edges. Let $T_{N}$ denote the space of trees of up to $N$ nodes, unique up to isomorphism. Then we have $G$'s associated uniform spanning tree distribution $P_G: T_N \rightarrow [0,1], \sum_{t \in T_N} P_G(t) = 1$, which can be sampled by Wilson's algorithm.

Now consider a second graph $H = (V_H, A_H)$, again of $N$ nodes and $M$ edges, with uniform spanning tree distribution $P_H: T_N \rightarrow [0,1], \sum_{t \in T_N} P_H(t) = 1$. Then my question is, does there exists such a $H$ such that,

$\forall t \in T_N, P_G(t) = P_H(t)$

and $G \not\simeq H$. That is, is the distribution $P_G$ unique up to graph isomorphism? Or does there exist a counter-example?

Or equivalently, can any two non-isomorphic directed graphs be distinguished by their uniform spanning tree distributions given sufficient (possibly exponentially many) samples?

Thanks for any thoughts and references!

Answer 1

The following two graphs are a counterexample:

For these graphs, not just the distribution but the multiset of spanning trees is the same, up to isomorphism. Both have the following $8$ spanning trees: