Given a (simple) graph $G=(V,E)$ with $V=\{1,...,n\}$ and let $A$ be its adjacency matrix.

I am interested in the representation theory (over $\Bbb R$) of the automorphism group $\def\Aut{\mathrm{Aut}}\Aut(G)$. One way to generate such representations is via spectral graph theory. If $\lambda$ is an eigenvalue of $A$ and $\{e_1,...,e_m\}$ is an orthonormal basis of the associated eigenspace $\def\Eig{\mathrm{Eig}}\Eig_\lambda(G)$, then the *rows* of the matrix

$$U:=\begin{pmatrix} | & & |\\ e_1 & \cdots & e_m \\ | & & | \end{pmatrix}$$

can be interpreted as the positions $v_i\in\Bbb R^m,i=1,...,n$ of the vertices of $G$ in a graph embedding. What is special about this embedding is, that it realizes all the symmetries of $G$. This means, for each automorphism $\phi\in\Aut(G)$, there is a linear map $M_\phi\in\mathrm{GL}(m,\Bbb R)$ with $v_{\phi(i)}=M_\phi v_i$. This gives a real representation $\Aut(G)\to\mathrm{GL}(m,\Bbb R),\phi\mapsto M_\phi$.

My questions are:

Was this construction of real representations of $\Aut(G)$ already studied somewhere in the literature?

and especially:

When are these respresentations (real)

irreducible?

This answer mentions graphs with trivial symmetry group but large eigenspaces, which therefore cannot provide irreducible representations. However, I am interested in graphs with a lot of symmetries, especially arc-transitive graphs. In all the cases I studied, all the representations turned out to be irreducible.

**Update**

- A similar question was asked on Math.StackExchange and received an interesting answer from C. Godsil. Especially the last parenthesized sentence leaves space for an interesting counter-example.
The searchable terminology seems to be "reducible/irreducible eigenspaces of graphs". At least this lead me to the following paper

G. Berkolaiko, W. Liu: Eigenspaces of Symmetric Graphs are not Typically Irreducible (2018)

However, I am not aware of a direct connection to the problem stated here, partially because the paper's terminology is not very familiar to me, yet.