distinct eigenvalues of adjacency matrix imply abelian automorphism group I'm trying to understand if my proof of the statement in the title is correct. I am in doubt since my argument implies a stronger conclusion. Here is my reasoning: 
Let $A$ be the adjacency matrix of a graph $G$ with the mentioned properties. The automorphism group of G is isomorphic to the group of permutation matrices That satisfy $\sigma A\sigma^t = A$. Since $A$ is symmetric, then by the spectral theorem it is unitarily diagonalizable, i.e, $A = UDU^t$ for some unitary matrix $U$. Then$$ \sigma UDU^t \sigma = UDU$$
$$ U^t \sigma U D U^t \sigma^t U = U^tUDU^tU = D.$$
since $U$ is unitary, $U^t (\sigma U)$ is also some permutation matrix. But since all entries of $diag(D)$ are distinct, it must be that $U^t\sigma U = I$. This means that the automorphism group is {1}.
I appreciate any help, thanks!
 A: Your proof is not correct. To pinpoint, the statement "since $U$ is unitary, .... is also a permutation matrix" is not correct.
You can check where your proof does not work considering the graph with two vertices and one edge ( a segment), with group of automorphisms $\mathbb{Z}/2$.
The correct proof is to see that if a matrix has all  the eigenvalues distinct then the group of invertible matrices that commute with it is abelian.  So the group of automorphisms, being a subgroup of that, is also abelian.
Why the above is true?  Say $A$ has all the eigenvalues distinct.  Then $A = M D M^{-1}$, where $D$ is diagonal with distinct elements. The group of matrices commuting with $D$ is $\frak{D}$, the group of diagonal invertible matrices, so abelian, and the group of matrices commuting with  $A$ is $M {\frak D} M^{-1}$, also abelian.
A: Cvetkovic, Rowlinson, Stanic and Yoon, Controllable graphs, available at http://elib.mi.sanu.ac.rs/files/journals/bltn/36/r6.pdf have a theorem that says the controllable graphs have a trivial automorphism group. A graph is called "controllable" if all its eigenvalues are distinct and "main". An eigenvalue is called main if its eigenspace contains a vector for which the sum of the coordinates is zero. 
The paper seems pretty readable, and has references to earlier results, which may amount to what you are asking about. 
