Expanding on Richard's comment: let me rename your graph to $S$ and consider the adjacency matrix $A$ abstractly as a linear operator acting on the free vector space $\mathbb{C}[S]$ on (the vertices of) $S$, and let $G$ be its automorphism group (this is why I wanted a new name). Then $\mathbb{C}[S]$ is a completely reducible representation of $G$ and $A$ is an endomorphism of this representation. Hence if we write

$$\mathbb{C}[S] \cong \bigoplus_i n_i V_i$$

where $V_i$ are the irreducibles, then $A$ is an element of the endomorphism algebra

$$\text{End}_G(\mathbb{C}[S]) \cong \prod_i M_{n_i}(\mathbb{C}).$$

This means more explicitly that $A$ is conjugate over $\mathbb{C}$ to a block diagonal matrix with a block for each isotypic component (hence its characteristic polynomial factors accordingly). In the nicest possible case the decomposition above is multiplicity-free in which case the endomorphism algebra is a product of copies of $\mathbb{C}$ and we just have that $A$ must act by a scalar $\lambda_i$ on each $V_i$ that occurs in the decomposition, which contributes a multiplicity of $\dim V_i$ to $\lambda_i$ as a root of the characteristic polynomial and hence, over $\mathbb{Q}$, contributes a multiplicity of $\dim V_i$ to the minimal polynomial of $\lambda_i$ as a factor of the characteristic polynomial.

(I think the result of this analysis comes out the same if you work over $\mathbb{Q}$ from the beginning but it's more annoying to describe.)

I work through a few examples of this in my old blog post The Schrodinger equation on a finite graph, where I was trying to understand via a toy model the quantum mechanical phenomenon of group symmetries introducing "degeneracies," which is physics speak for eigenvalues (of the Hamiltonian in this case) of multiplicity greater than $1$.

The "most degenerate" case is the complete graph $S = K_n$, where $G = S_n$ and the corresponding representation is a copy of the trivial representation and an irreducible representation of degree $n-1$. This means the adjacency matrix $A$ must have at most two eigenvalues, one with multiplicity $1$ and one with multiplicity $n-1$, which turn out to be $n-1$ and $-1$ respectively (this is easily computed by computing $\text{tr}(A)$ and $\text{tr}(A^2)$, or just finding all the eigenvectors of $A + I$), inducing a factorization

$$\det (tI - A) = (t - n + 1)(t + 1)^{n-1}$$

with a factor of multiplicity $n-1$.

One of the "least degenerate" cases where the automorphism group still acts transitively on vertices is $S = C_n$ the cycle graph, where $G = D_n$ is the dihedral group and the corresponding representation splits up into mostly $2$-dimensional irreps. This reflects the fairly mild degeneracies of the eigenvalues of the adjacency matrix, which are $2 \cos \frac{2\pi k}{n}, k = 0, \dots n-1$, and/but which also organize themselves into nontrivial Galois orbits coming from the action of the Galois group of $\mathbb{Q}(\zeta_n)$.

three, I would probably try to get some intiution for its structure by plotting the 3-dimensional spectral graph realizations to these eigenvalues. Maybe some of these are of a quite recognizable shape (a Platonic solid etc). See here for what I mean by "spectral realization". $\endgroup$