TITLE: Are there only finitely many distinct cubic walk-regular graphs that are neither vertex-transitive nor distance-regular?
QUESTION [7 upvotes]: The class of walk-regular graphs contains the vertex-transitive graphs and the distance-regular graphs. However, there are walk-regular graphs that are neither vertex-transitive nor distance-regular. In particular, I believe that I found a first example of a cubic walk-regular graph that is neither vertex-transitive nor distance-regular on 20 vertices: its Graph6 code is SsP@@?OC?S@C@_@C?K?A_?AG?D??C_??]. 
My question here is: are there only finitely many cubic walk-regular graphs (up to isomorphism) that are neither vertex-transitive nor distance-regular (or is it e.g. possible to construct an infinite family of graphs of this kind)?
There are e.g. only finitely many distinct cubic distance-regular graphs  - could it e.g. also be true that there are only finitely many distinct non-vertex-transitive cubic walk-regular graphs?

REPLY [6 votes]: What follows is a proof that semisymmetric graphs are walk-regular.
Say vertices $u$ and $v$ in a graph $X$ are cospectral if the graphs $X\setminus u$ and $X\setminus v$ are cospectral. If $X$ is semisymmetric, then the vertices in each colour class are cospectral. Two vertices $u$ and $v$ in a bipartite graph are cospectral if, for each non-negative integer $k$, the number of closed walks of length $2k$ starting at $u$ equals the number starting at $v$. [This is proved, for example, in Section 8.13 in one of my favourite texts on algebraic graph theory. :-)]
The adjacency matrix $A$ of any bipartite graph can be written in the form
\[
  A =\begin{pmatrix}0&B\\ B^T&0\end{pmatrix}
\]
and hence
\[ 
  A^{2k} = \begin{pmatrix}(BB^T)^k&0\\ 0&(B^TB)^k&\end{pmatrix}
\]
Since $X$ is semisymmetric, the diagonal entries of $(BB^T)^k$ are all equal, as are the diagonal entries of $(B^TB)^k$. 
By the cyclic symmetry of trace, we see that $(BB^T)^k$ and $(B^TB)^k$ have the same trace. Therefore the diagonal entries of these two matrices are equal, and it follows that $X$ is walk regular.