Let $X$ be a separable Hilbert space, $D$ nonnegative definite (by which I also mean self-adjoint) and trace class operator on $X$. Let $A$ be a compact and injective operator with dense range $R$, so that $A^{-1}$ is densely defined. Define $B=ADA^{-1}$ on $R$ and assume that $B$ is bounded. Then it can be extended to the closure of $R$, which is $X$. Assume that $B$ is also trace class.
Question: is it true that trace$(B)=$trace$(D)$?
Here are some thoughts.
(1) This is true in the finite-dimensional case, of course.
(2) This is true if $A$ is trace class and $DA^{-1}$ is bounded.
(3) This is true if $A$ and $D$ commute: $AD=DA$.
(4) Let $(v_k)$ be an basis of $X$ with $Dv_k=\lambda_kv_k$, and let $u_k=Av_k$. Then $Bu_k=\lambda_k u_k$, so $(u_k)$ are eigenvectors of $B$ but they are not orthogonal. Heuristically, these are the only eigenvectors of $B$ (if $Bv=\lambda v$, then $DA^{-1}v=\lambda A^{-1}v$; this is not a proof, of course). So the question is somehow asking: does the trace equal the sum of eigenvalues even if the eigenvectors are not orthogonal?
(5) I do not mind assuming things about the rate of decay of the eigenvalues of $D$ or the singular values of $A$ if it helps to get a positive answer. There is also some more structure in my particular case, but I don't think that it is relevant: in my case $A$ is a product of positive definite operators, one is Hilbert--Schmidt and the other bounded.
Since $A$ is compact, $A^{-1}$ is bounded by the bounded inverse theorem. Then, since $D$ is trace class, $DA^{-1}$ is also trace class. Finally, since $A$ is compact, the result follows $\mathrm{tr}[(A)(DA^{-1})]=\mathrm{tr}[(DA^{-1})(A)]=\mathrm{tr}(D)$.