It is well known that


1) if there exists a non-trivial automorphism of a graph $G$ with corresponding permutation matrix $P$ then if $(v,\lambda)$ is an eigenvector-eigenvalue pair of the graph Laplacian $L(G)$ then $(Pv,\lambda)$ is also an eigenvector-eigenvalue pair (if $v$ and $Pv$ are linearly independent then this gives rise to eigenvalues with multiplicity greater than 1) and, 


2) if all the eigenvalues of the $L(G)$ are simple than every automorphism of G has order 1 or 2. 


If $G$ exhibits only a trivial automorphism ($G$ is asymmetric) can it be said that $L(G)$ has no repeated eigenvalues?


If not, a counterexample would be most helpful. I haven't found one on a brute force check of all graphs up to 9 nodes.


(I am assuming unweighted, undirected graphs)



Does graph asymmetry imply all eigenvalues of the graph Laplacian are simple?
This is false.


There are strongly regular graphs with trivial automorphism group; these will have many repeated eigenvalues (both for adjacency matrix and Laplacian).


You can find some examples in the answers to this question:


[Are "almost all" strongly regular graphs rigid?](https://mathoverflow.net/questions/41194/are-almost-all-strongly-regular-graphs-rigid)


