Suppose that I have $n$ matrices $A_1, \ldots, A_n \in \mathbb{R}^{m \times m}$. Can I find $n$ new matrices $B_1, \ldots, B_n \in \mathbb{R}^{n \times n}$ that have the same 3-way cyclic traces: $$ \forall i, j, k : \mathrm{Tr}(A_i A_j A_k) = \mathrm{Tr}(B_i B_j B_k)?$$ By analogy, if I had $n$ vectors $v_1, \ldots, v_n \in \mathbb{R}^m$, it would be easy to construct new vectors $u_1, \ldots, u_n \in \mathbb{R}^n$ that have the same inner products (by choosing an orthonormal basis for the span of the $v_i$ and then writing each $v_i$ in that basis). Parameter counting suggests there should be matrices $B$ that match a given set of cyclic traces (we have $n^3$ parameters to pick and only $n^3/3$ constraints), but I have no idea how you could pick them "naturally" and don't have any reason beyond parameter counting to think they exist.

Context: I'm interested in succinctly summarizing the interactions between the matrices $A$ in a way that lets me make reasonable guesses about arbitrary cyclic traces like $\mathrm{Tr}(A_i A_j A_k A_l)$. I've found no leads on how to guess higher-order traces given lower-order traces, finding a small set of matrices with a given set of low-order traces seems like a natural first step on the problem if it's possible. (I considered instead assuming that higher free cumulants are zero, but this will tend to produce unreasonable estimates including $\mathrm{Tr}(X^2) < 0$.)