Let $\rho = \rho(x)$ be a $2\times2$ matrix (don't know if it is necessary, but $\rho$ is a density operator) and $I$ be the (two-dimensional) identity matrix. I have two matrices $A$ and $B$, where $A = \rho^{\otimes j}\otimes \mathrm{I}^{N-j}$ and $B$ consisting of some permutation of the tensor products with the same amount of $\rho$ and $I$ terms, e.g. $B = \rho^{\otimes j-1}\otimes \mathrm{I}\otimes \rho \otimes \mathrm{I}^{N-j-1}$. Is it in general true then that the eigenvalues of $\int_0^g\mathrm{d}xA~f(x)$ are equal to the eigenvalues of $\int_0^g\mathrm{d}xB~f(x)$ for all $g$ and all (well-behaved) functions $f(x)$? For my specific forms of $\rho$ and $f(x)$ it seems to hold by numerical calculation, and it also seems logical for it to hold, but there must also be some way to see this formally/intuitively.
Since $A$ and $B$ are permutation similar, a more general question could be whether or not for any two permutation similar matrices $C(x)$ and $D(x)$ we have that the eigenvalues of $\int_0^g\mathrm{d}xC(x)~f(x)$ are always equal to the eigenvalues of $\int_0^g\mathrm{d}xD(x)~f(x)$.