From Wikipedia.
In linear algebra, two n-by-n matrices A and B are called similar if
$$ B = P^{-1} A P$$
for some invertible n-by-n matrix $P$.
If $P$ is a permutation matrix, $A$ and $B$ are permutation similar.
Two graphs $G,H$ are isomorphic iff their adjacency matrices
$A_G,A_H$ are permutation similar.
If $A_G$ and $A_H$ are not similar then $G$ and $H$ are not isomorphic.
It is possible $A_G$ and $A_H$ to be similar and $G$ and $H$ are not isomorphic.
Experimentally this doesn't happen often.
Recognizing matrix similarity is polynomial, $O(n^3)$.
$A$ and $B$ are similar if and only if they have the same rational canonical form (AKA Frobenius form).
Is there characterization of the graphs for which
$A_G \sim A_H \implies G \cong H$?
Consider the following algorithm for graph isomorphism.
Recursively delete vertex from $G$ and from $H$ and
check the matrices for similarity. If we don't hit
bad case we will find isomorphism in polynomial time.
The exponential case is if we hit bad cases very often.
Is there construction of non-isomorphic graphs for
which the above algorithm is exponential?
Added
The sequence of non-similar matrices of graphs of order $n$
starts
1 , 2 , 4 , 11 , 33 , 151 , 988 , 11453 , 247357
This coincides with OEIS A082104 Number of distinct characteristic polynomials among all simple undirected graphs on n nodes
If true, this would imply that symmetric matrices with $0,1$ entries
are similar iff their characteristic polynomials are equal
(which is likely known).
Best Answer
There is no characterization known of when a graph is determined by its spectrum. The probability that a tree on $n$ is determined by its characteristic polynomial goes to zero as $n$ tends to infinity. It is conjectured that for graphs the probability goes to one.