If $A$ has conjugate eigenvalues then it has in some basis a representation $PAP^{-1}=A_a=\begin{bmatrix} a & -b \\ b & a \end{bmatrix} $.

Let $B_a$ be any matrix $\begin{bmatrix} c & d \\ e & f \end{bmatrix} $ which is commutative with $A_a$ in this basis

i.e. $A_aB_a=B_aA_a$.

Compare
$\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} c & d \\ e & f \end{bmatrix} =
\begin{bmatrix} c & d \\ e & f \end{bmatrix}
\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$

and you should obtain conditions for $c,d,e,f$

If the representation is of the form like $A_a$ ( $c=f$ and $d=-e$) then you'll achieve the desired result because eigenvalues are constant under a change of basis (so their property of conjugation) and commutation for both matrices holds also under a change of basis.