Diagonalization of a matrix given by a finite group

Question

$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\GL}{\mathrm{GL}} \newcommand{\Gal}{\mathrm{Gal}} \newcommand{\matr}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix}} $

Let $G$ be a finite group and $S : G \to \C$ be any map. Define the matrix $$ M = (S(xy^{-1}))_{x, y \in G} $$

Is it true that $M$ is diagonalizable of $\C$ ? Note: it would be equivalent to take $M' := (S(xy))_{x, y \in G} $.

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Thoughts: It is true if $G = \Z / g \Z$ is a cyclic group of order $g>1$, as one can check that for $U := (e^{2 \pi i x y / g})_{1 \leq x, y \leq g}$, we have $MU = UD$ for some diagonal matrix $D$. Moreover, $U$ is invertible (check the SAGE code below). This was in some paper by Takashi Ono. Apparently, it also works for some non-abelian groups, but I don't know how to proceed.

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UCF = UniversalCyclotomicField();
for g in range(1, 6):
    U = matrix(UCF, [[E(g)^(x*y) for x in range(g)] for y in range(g)])
    print(g, U.determinant())#, CC(U.determinant()))
    print(U, "\n")
    print(U.inverse(), " ====== \n\n")