Let $Q$ be unitary and $M$ be Hermitian positive semidefinite (psd), so that $QMQ^*$ is also Hermitian psd. What conditions (probably on $f(M)$) makes the trace inequality $$\operatorname{Tr}\Big(QMQ^*f(M)\Big) \le \operatorname{Tr}\Big(QQ^*Mf(M)\Big) = \operatorname{Tr}\Big(Mf(M)\Big)$$ true? I searched and I'm still searching.
If it helps, I know that $\operatorname{Tr}f(M)\ge 0$.
Thank you.
In general, if $X$ is Hermitian and $\Sigma$ is a nonnegative diagonal matrix whose diagonal entries are arranged in decreasing order, then $\operatorname{Tr}(Q\Sigma Q^\ast X)$ is maximised over $U(n)$ when $Q^\ast XQ$ is also a diagonal matrix whose diagonal entries are arranged in decreasing order. (E.g. see my other answer to a more general problem.)
It follows that if $f$ is any analytic function such that $f(\mathbb R)\subseteq\mathbb R$ (note: $f$ can take negative values; $f(\mathbb R)$ isn't required to lie inside the nonnegative real line) and $f$ is non-decreasing on $\mathbb R$, then $M$ and $f(M)$ can be simultaneously unitarily diagonalised in a way that their diagonal entries are arranged in decreasing order and hence $\operatorname{Tr}(QMQ^\ast f(M))\le\operatorname{Tr}(Mf(M))$ for every positive semidefinite $M$.