$\newcommand{\End}{\operatorname{End}}\newcommand{\hdual}{\mathsf H}$Let $V$ be a finite dimensional vector space and $\Phi \in \End(V)$.
If $\hat\Phi,\Psi \in \End(V)$ such that $\hat\Phi$ is an isometry, $\Psi$ is positive semi-definite, and $\Phi = \hat\Phi \Psi$, then $\Psi = \sqrt{\Phi^\hdual\Phi}$, where $\sqrt{\cdot}$ is the unique positive semi-definite square root of a PSD operator.
How to prove this theorem?
My attempt: $\Phi$ has a polar decomposition $\Phi=\Theta\sqrt{\Phi^\hdual\Phi}$, where $\Theta$ is isometric. Hence $$\Psi = \hat\Phi^{-1} \circ \Theta \circ \sqrt{\Phi^\hdual\Phi}$$ Every isometry is bijective and the combination of isometries is still isometric. The proof would be complete if we can show that a positive semi-define operator combined with an isometry is PSD iff the isometry is identity, but I do not know how to complete that.
If you already know that a positive semi-definite operator has a unique positive semi-definite root then your questions amounts to the following calculation:
$$ \Phi^{H} \Phi = (\hat{\Phi} \Psi)^{H} (\hat{\Phi} \Psi) = \Psi^H \hat{\Phi}^H \hat{\Phi} \Psi = \Psi^H \Psi = \Psi^2. $$
This shows that $\Psi$ is a positive semi-definite square root of the positive semi-definite operator $\Phi^H \Phi$ and so by uniqueness $\Psi = \sqrt{\Phi^{H} \Phi}$.