Let $A(x,t),B(x,t)$ be matrix-valued functions that are independent of $\xi=x-t$ and satisfy $$A_t-B_x+AB-BA=0$$ where $X_q\equiv \frac{\partial X}{\partial q}$.
Why does it then follow that $$\frac{d }{d \eta}\textrm{Trace}[(A-B)^n]=0$$ where $n\in \mathbb N$ and $\eta=x+t$?
Is there a neat way to see that this is true?
(This may not be a neat way to prove the assertion, but it's a proof anyway.) Let $\eta=x+t$ and $\nu=x-t$. Then $x=\eta+\nu$ and $t=\eta-\nu$ are functions of $\eta$ and $\nu$, $A=A(\eta+\nu,\eta-\nu)$ and similarly for $B$. As both $A$ and $B$ are independent of $\nu$, by the total derivative formula, we get \begin{align*} 0 = \frac{dA}{d\nu} &= \frac{\partial A}{\partial x} - \frac{\partial A}{\partial t},\\ 0 = \frac{dB}{d\nu} &= \frac{\partial B}{\partial x} - \frac{\partial B}{\partial t} \end{align*} and hence \begin{align*} \frac{dA}{d\eta} &= \frac{\partial A}{\partial x} + \frac{\partial A}{\partial t} = 2\frac{\partial A}{\partial t},\\ \frac{dB}{d\eta} &= \frac{\partial B}{\partial x} + \frac{\partial B}{\partial t} = 2\frac{\partial B}{\partial x}. \end{align*} Therefore $$ \frac{d(A-B)}{d\eta} = 2\left(\frac{\partial A}{\partial t} - \frac{\partial B}{\partial x}\right) = 2(BA - AB). $$ and \begin{align*} \frac{d}{d\eta}\operatorname{trace}[(A-B)^n] &= n \operatorname{trace}\left[(A-B)^{n-1}\frac{d(A-B)}{d\eta}\right]\\ &= 2n \operatorname{trace}[(A-B)^{n-1} (BA-AB)].\tag{1} \end{align*} Let $\mathcal{P}_m$ denotes the set of products of $m$ matrices from $\{A,B\}$ (e.g. $\mathcal{P}_2=\{AA,AB,BA,BB\}$). Then the function $f:\mathcal{P}_m\to \mathcal{P}_m$ defined by \begin{cases} f(B^m) = B^m,\\ f(pAB^k) = B^kAp &\text{ for all } 0\le k<m \text{ and } p\in\mathcal{P}_{m-k-1} \end{cases} is a bijection. Since $\operatorname{trace}(AqB)=\operatorname{trace}(Bf(q)A)$ for all $q\in\mathcal{P}_{n-1}$ and the degree of $B$ is preserved by $f$, it follows that $\operatorname{trace}[A(A-B)^{n-1}B]=\operatorname{trace}[B(A-B)^{n-1}A]$. Consequently, $$\operatorname{trace}[(A-B)^{n-1} (BA-AB)]=0$$ for all square matrices $A,B$ and $(1)$ evaluates to zero.