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Let G be a graph. If G is simple(i.e no loops), and the eigenvalues of adjacency matrix A are distinct, then the automorphism of G is abelian. It seems that the automorphism from G to G itself is only permutation upon the vertex with incidence function composing with this permutation. But I do not know how to show this permutation(cyclic group) is the whole automorphism. Any hints will be helpful.

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There are some details for you to fill in, but here is an outline argument that seems to work. The automorphism group of $G$ is isomorphic to the set of permutation matrices $P$ such that $P^{-1}AP = A$; i.e., the permutation matrices that commute with $A$. Now, $A$ is similar to a diagonal matrix $D$, with the (distinct) eigenvalues on the diagonal. The (invertible) matrices that commute with $D$ must be themselves diagonal (because of the distinct eigenvalues), and therefore commute with one another. Since these are conjugate to the matrices in the (copy of the) automorphism group, the automorphism group must itself be commutative.

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