= Solution
The relation is
$$
A_{ij}=\alpha\delta_{ij}B_{kk}+\beta B_{ij}+\gamma B_{ji}.
$$
Taking its trace, symmetric traceless part, and antisymmetric part separately gives
$$
B_{kk}=\frac{A_{kk}}{3\alpha+\beta+\gamma},\qquad
B_{(ij)}-\frac13\delta_{ij}B_{kk}
=\frac{A_{(ij)}-\frac13\delta_{ij}A_{kk}}{\beta+\gamma},
\qquad
B_{[ij]}=\frac{A_{[ij]}}{\beta-\gamma}.
$$
Equivalently,
$$
\boxed{
B_{ij}=
\frac{A_{ij}+A_{ji}}{2(\beta+\gamma)}
+\frac{A_{ij}-A_{ji}}{2(\beta-\gamma)}
-\frac{\alpha\delta_{ij}A_{kk}}{(\beta+\gamma)(3\alpha+\beta+\gamma)}.}
$$
If $\beta=-\gamma\ne0$, $\alpha\ne0$, and $A_{ij}=0$, its symmetric part forces $B_{kk}=0$ and its antisymmetric part forces $B_{[ij]}=0$. Thus \b[the solutions are exactly the symmetric traceless tensors $B_{ij}$].
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