 Solution

For a positive operator $A$, the spectral theorem writes $A={\sum }_j{a}_j|j⟩⟨j|$ with $a_j\ge 0$. Hence its unique positive square root of an operator is $\sqrt{A}={\sum }_j\sqrt{a_j}|j⟩⟨j|$. Conjugating this spectral decomposition shows $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$ for a unitary operator $U$.

Let $\Delta ={\rho }_1-{\rho }_2$ and ${\rho }_i^{\prime }=U{\rho }_i{U}^{†}$. Then ${\Delta }^{\prime }=U\Delta U^{†}$ and $({\Delta }^{\prime }{)}^{†}{\Delta }^{\prime }=U{\Delta }^{†}\Delta U^{†}$. The square-root covariance and the cyclic property of the trace giveFor the quantum fidelity, covariance first gives $({\rho }_1^{\prime }{)}^{1/2}=U{\rho }_1^{1/2}{U}^{†}$. ThereforeApply the same square-root identity and the cyclic property of the trace to obtainThus unitary invariance of trace distance and fidelity givesThe same $U$ acts on both quantum states: these quantities depend on the relative distinguishability of the quantum states, not on a choice of basis.