= Solution
The stated coordinate substitution splits the polynomial <basis> into two three-cycles:
$$
x^2\longmapsto y^2\longmapsto z^2\longmapsto x^2,\qquad
xy\longmapsto yz\longmapsto xz\longmapsto xy.
$$
The identity fixes all six <basis vectors>; neither nonidentity element fixes any. A <permutation representation> has <character> equal to the number of fixed <basis vectors>, hence
$$
\boxed{\chi_{\widetilde R}(0)=6,\qquad
\chi_{\widetilde R}(1)=\chi_{\widetilde R}(2)=0.}
$$
The <representation multiplicity> of each <irreducible character> is therefore
$$
m_k=\langle\chi_k,\chi_{\widetilde R}\rangle=\frac13\cdot6=2,
\qquad
\boxed{\widetilde R\cong2\rho_0\oplus2\rho_1\oplus2\rho_2.}
$$
An explicit <eigenbasis> confirms the result. For $k=0,1,2$, put
$$
f_k=x^2+\alpha^{-k}y^2+\alpha^{-2k}z^2,\qquad
h_k=xy+\alpha^{-k}yz+\alpha^{-2k}xz.
$$
Then $\widetilde R(1)f_k=\alpha^kf_k$ and $\widetilde R(1)h_k=\alpha^kh_k$. Thus each three-cycle carries one copy of every <irreducible representation>. These signs use precisely the source's substitution action; taking an inverse substitution would interchange $\rho_1$ and $\rho_2$, leaving the total decomposition unchanged.
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