= Solution
Right multiplication is linear because $(sX+tY)A=sXA+tYA$. In the supplied row-entry <basis>, write $X=\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix}$. Multiplying on the right gives coordinates $(ax_1+cx_2,bx_1+dx_2,ax_3+cx_4,bx_3+dx_4)^T$. Therefore
$$
\boxed{[\rho_A]=\begin{pmatrix}a&c&0&0\\b&d&0&0\\0&0&a&c\\0&0&b&d\end{pmatrix}=\operatorname{diag}(A^T,A^T).}
$$
The transpose is ordinary, with no complex conjugation. Block determinants and invariance of determinant under transpose prove
$$
\boxed{\chi_{\rho_A}(t)=\det(tI_2-A^T)^2=\chi_A(t)^2.}
$$
For every polynomial $p$, powers of right multiplication give $p(\rho_A)(X)=Xp(A)$. If $p(A)=0$ then $p(\rho_A)=0$; if $p(\rho_A)=0$, take $X=I$ to infer $p(A)=0$. They have identical annihilating polynomials, so their unique monic least-degree annihilator, the <minimal polynomial>, is the same. Thus \b[$m_{\rho_A}=m_A$], as in the <characteristic and minimal polynomials of right multiplication> result.
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