Let $A_1,\ldots,A_m,B_1,\ldots,B_m\in\Bbb C^{n\times n}$ be matrices such that $$ 0 = \sum_{i=1}^m\operatorname{trace}( A_i^{\mathrm T} \cdot SB_iS^{-1}) = \operatorname{trace}( A_1^{\mathrm T} SB_1S^{-1} + \cdots + A_m^{\mathrm T} SB_mS^{-1} ) $$ for all $S\in\operatorname{GL}_n(\Bbb C)$.
Furthermore, assume that the $A_i$ are linearly independent and traceless, i.e. $\operatorname{trace}(A_i)=0$. The exercise is to show that all $B_i$ are scalar matrices.
Another way to say this is, using the scalar product $\langle A,B\rangle :=\operatorname{trace}(A^{\mathrm T} B)$: The $A_i$ are all orthogonal to the identity matrix $I$ and we have $\sum_{i=1}^m \langle A_i, SB_iS^{-1} \rangle = 0$ for all $S\in\operatorname{GL}_n$. Prove that the $B_i$ must be multiples of $I$.
I am reasonably sure this is correct, but counterexamples are also welcome.
Edit: I had a faulty attempt here to prove the $m=1$ case, I will try to fix it.