Let $A,B$ be linear, compact, self-adjoint and even trace-class operators.
Can I bound $\|ABA^{-1}\|_{HS}$ by the norm of $\|B\|_{HS}$ somehow? (Where the bound does not depend on $\|A^{-1}\|$.)
Here $A^{-1}$ means the pseudo inverse i.e. for $$Af=\sum_{j}\lambda_j\langle f,e_j\rangle e_j$$ we set $A^{-1}$: $$ A^{-1}f=\sum_{j}\lambda^{-1}_j\langle f,e_j\rangle e_j. $$
In general one cannot even guarantee that $ABA^{-1}$ is a Hilbert-Schmidt operator (or a bounded operator for that matter) as the following example will illustrate. Let \begin{align*} A&:=\sum_{n=1}^\infty\frac1{n!}|e_n\rangle\langle e_n|\\ B&:=\sum_{n=1}^\infty\frac1{n^{3/2}}\big(|e_n\rangle\langle e_{n+1}|+|e_{n+1}\rangle\langle e_{n}|\big) \end{align*} be operators on $\ell^2(\mathbb N)$ with $(e_n)_{n=1}^\infty$ the standard basis. Obviously, $A$ is self-adjoint and trace class, and $B$ is self-adjoint. Let us quickly see that $B$ is also trace class; it suffices to show that $\sum_{n=1}^\infty|\langle\phi_n,B\phi_n\rangle|<\infty$ for all orthonormal bases $(\phi_n)_{n=1}^\infty$ of $\ell^2(\mathbb N)$. Indeed, we compute \begin{align*} \sum_{n=1}^\infty|\langle\phi_n,B\phi_n\rangle|&\leq\sum_{n=1}^\infty\sum_{p=1}^\infty \frac1{p^{3/2}}\big|2\operatorname{Re}\big(\langle\phi_n,e_p\rangle\langle e_{p+1},\phi_n\rangle\big)\big|\\ &\leq\sum_{n=1}^\infty\sum_{p=1}^\infty 2\frac1{p^{3/2}}\big|\langle\phi_n,e_p\rangle\langle e_{p+1},\phi_n\rangle\big|\\ &=2\sum_{p=1}^\infty \frac1{p^{3/2}}\sum_{n=1}^\infty\big|\langle\phi_n,e_p\rangle\langle e_{p+1},\phi_n\rangle\big|\\ &\leq 2\sum_{p=1}^\infty \frac1{p^{3/2}}\Big(\sum_{n=1}^\infty\big|\langle\phi_n,e_p\rangle|^2\Big)^{1/2}\Big(\sum_{n=1}^\infty\langle e_{p+1},\phi_n\rangle\big|\Big)^{1/2}=2\sum_{p=1}^\infty\frac1{p^{3/2}}\,. \end{align*} Here, in the second-to-last line we used a version of Fubini's theorem to interchange summation (because all summands are non-negative), and in the last line we used the Cauchy-Schwarz inequality on $\ell^2(\mathbb N)$ as well as Parseval's identity (because $\|e_p\|=1$). But the upper bound we found is well known to converge so we showed that $B$ is trace class.
A straightforward formal computation shows $$ ABA^{-1}=\sum_{n=1}^\infty \frac1{\sqrt n}|e_n\rangle\langle e_{n+1}|+\frac1{(n+1)n^{3/2}}|e_{n+1}\rangle\langle e_n| $$ so \begin{align*} \|ABA^{-1}\|_\mathrm{HS}^2=\sum_{j,k=1}^\infty|\langle e_j,ABA^{-1}e_k\rangle|^2&=\sum_{n=1}^\infty\Big(\frac1{\sqrt n}\Big)^2+\sum_{n=1}^\infty\Big(\frac1{(n+1)n^{3/2}}\Big)^2\\ &=\underbrace{\sum_{n=1}^\infty \frac1n}_{=\infty}+\underbrace{\sum_{n=1}^\infty\frac1{(n+1)^2n^3}}_{<\infty} \end{align*} meaning that $ABA^{-1}$ is not a Hilbert-Schmidt operator (1, 2), although both $A,B$ are (even trace class).