= Solution
For <permutations> $\sigma,\tau$, the prescribed <linear maps> satisfy
$$
R(\sigma)R(\tau)e_k=R(\sigma)e_{\tau(k)}=e_{\sigma(\tau(k))}=R(\sigma\tau)e_k.
$$
Since this holds on a <basis>, $R(\sigma)R(\tau)=R(\sigma\tau)$. Also $R(e)=I$ and $R(\sigma)^{-1}=R(\sigma^{-1})$. Hence $R$ is a <representation>.
The <trace> of a <permutation matrix> counts its fixed coordinate <basis vectors>. There are respectively $3,1,0$ fixed vectors on the three <conjugacy classes>, so $\chi_R=(3,1,0)=\chi_{\mathbf1}+\chi_D$. Consequently
$$
\boxed{R=\mathbf1\oplus D.}
$$
In particular the <sign representation> is absent.
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