Let $U$ be a unitary operator on a Hilbert space $H$ and $Q \in B(H)$ is positive and invertible, is it true that if $\|QUQ^{-1}\| = 1$, then $Q$ commutes with $U$?

One can show that this is the case if the spectrum of $Q$ is finite, but I don't know how to treat the general case. Note that one always has $\| Q U Q^{-1} \| \ge 1$ because $Q U Q^{-1}$ and $U$ have the same spectrum, and one only needs to show that if the norm is $1$, then $Q U Q^{-1}$ must be unitary then conclude by the uniqueness of the left and right polar decomposition.

Edit. This is solved in the negative by Ozawa in the comments. For the conclusion to hold, one indeed needs the positive operator to have finite spectrum.

Let $U$ be a unitary operator on a Hilbert space $H$ and $Q \in B(H)$ is positive and invertible, is it true that if $\|QUQ^{-1}\| = 1$, then $Q$ commutes with $U$?

One can show that this is the case if the spectrum of $Q$ is finite, but I don't know how to treat the general case. Note that one always has $\| Q U Q^{-1} \| \ge 1$ because $Q U Q^{-1}$ and $U$ have the same spectrum, and one only needs to show that if the norm is $1$, then $Q U Q^{-1}$ must be unitary then conclude by the uniqueness of the left and right polar decomposition.

Edit. This is solved in the negative by Ozawa in the comments. For the conclusion to hold, one indeed needs the positive operator to have finite spectrum.

Edit 2. Now come to think about it, this seems to only hold in finite dimensional case by Hadamard's inequality, Ozawa's example even can be used to show that this does not hold whether the spectrum of $Q$ is finite or not. Seems that the reason this can fail really lies in the infinite dimensionality.

Let $U$ be a unitary operator on a Hilbert space $H$ and $Q \in B(H)$ is positive and invertible, is it true that if $\|QUQ^{-1}\| = 1$, then $Q$ commutes with $U$?

One can show that this is the case if the spectrum of $Q$ is finite, but I don't know how to treat the general case. Note that one always has $\| Q U Q^{-1} \| \ge 1$ because $Q U Q^{-1}$ and $U$ have the same spectrum, and one only needs to show that if the norm is $1$, then $Q U Q^{-1}$ must be unitary then conclude by the uniqueness of the left and right polar decomposition.

Let $U$ be a unitary operator on a Hilbert space $H$ and $Q \in B(H)$ is positive and invertible, is it true that if $\|QUQ^{-1}\| = 1$, then $Q$ commutes with $U$?

One can show that this is the case if the spectrum of $Q$ is finite, but I don't know how to treat the general case. Note that one always has $\| Q U Q^{-1} \| \ge 1$ because $Q U Q^{-1}$ and $U$ have the same spectrum, and one only needs to show that if the norm is $1$, then $Q U Q^{-1}$ must be unitary then conclude by the uniqueness of the left and right polar decomposition.

Edit. This is solved in the negative by Ozawa in the comments. For the conclusion to hold, one indeed needs the positive operator to have finite spectrum.