It's well known that for finite-dimensional linear operators $A$ and $B$, $\mathrm{tr}(AB) = \mathrm{tr}(BA)$, from which it follows that the trace of a product of multiple matrices is left invariant under arbitrary cyclic permutations.
Furthermore, we have that $$\mathrm{tr}\left[ (AB)^n \right] = \mathrm{tr} [ \underbrace{AB AB \dots AB}_{\text{$n$ times}} ] = \mathrm{tr}[ \underbrace{BA BA \dots BA}_{\text{$n$ times}}] = \mathrm{tr}\left[ (BA)^n \right],$$
and so by linearity we should have than $\mathrm{tr}[f(AB)] = \mathrm{tr}[f(BA)]$ for any analytic function $f$, as long as both arguments lie within the domain of convergence. And more generally, this result should extend to arbitrary cyclic permutations of the matrices in the argument of $f$.
- Is it true that $\mathrm{tr}[f(ABC\dots)]$ is invariant under any cyclic permutation of the matrices in the argument of $f$, as long as both matrix products lie within the domain of convergence of the Maclaurin series for $f$?
- Is it possible for a cyclic permutation of the matrices to move the product in or out of the domain of convergence?
- Does the story change if we consider infinite-dimensional trace-class linear operators instead of finite-dimensional ones?
$AB$ and $BA$ always have the same nonzero eigenvalues, and these should determine the convergence of the series for $f$ at $AB$ and $BA$.