Does graph asymmetry imply all eigenvalues of the graph Laplacian are simple?

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)

Take two asymmetric $d$-regular graphs $H_1,H_2$, and let $G$ be their disjoint union. Then $d$ will be a repeated eigenvalue.
If you want $G$ connected, take the complement of the graph obtained by the above construction. Graph complements preserve asymmetry and repeated eigenvalues.