**Setting** Let $G=(V,E)$ be an undirected graph. A *walk* $\pi$ in $G$ of length $k$ is a sequence of $k+1$ vertices $v_1,\ldots,v_{k+1}$ such that for each $i\in[1,k]$,
$\{v_i,v_{i+1}\}\in E$. Let $H=(W,F)$ be another undirected graph having the same number of vertices as $G$, i.e., $|V|=|W|=n$.

If for each $k$, $G$ and $H$ have the same number of walks of length $k$, then it is known that there is matrix $Q$ such that $A_G\cdot Q=Q\cdot A_H$, where $A_G$ and $A_H$ denote adjacency matrices of $G$ and $H$, respectively, and such that $Q\cdot\mathbf{1}=\mathbf{1}$ and $\mathbf{1}^t\cdot Q=\mathbf{1}^t$, where $\mathbf{1}$ is the $n\times 1$-vector consisting of all ones. (A matrix with this property is sometimes called doubly quasi-stochastic). The converse also holds, i.e., when $A_G\cdot Q=Q\cdot A_H$ holds for a doubly quasi-stochastic matrix, then for any $k$, $G$ and $H$ have the same number of walks of length $k$.

**Question** Let us consider the *directed graph* (digraph) case. Is there an example of two digraphs with the same number of vertices that have same number of walks of length $k$, for any $k$, yet there is **no** doubly quasi-stochastic matrix $Q$ such that $A_G\cdot Q=Q\cdot A_H$?