The original question is Here(In polar decomposition(A=RU) decomposition, How to extend Isometry into whole complex Euclidean space)
But I make brief conditions that we need.
Let $X$ is complex Euclidean space and $A$ is linear mapping of $X$
Conditions
i) $A$ is not invertible,
ii) $AA^*=R^2$ where $AA^*$, $R$ is both nonnegative self-adjoint mapping
iii) $\parallel Rx\parallel = \parallel A^*x\parallel$ for any $x\in X$
iv) If $Rx=Ry$, $A^*x=A^*y$
For any $u$ in the range of $R$, $u=Rx$, we can define $Vu$ as $A^*x$
By upper condition, $V$ is isometry mapping of $Range(R)$ to $Range(A^*)$
---Here, I can't extend $V$ to whole space, $X$
I tried with following steps.
For any $x\in X$, $x$ can be decomposed by $x=u+v$ where $u\in Range(R), (u,v)=0$
Let $u=Ra$ for some $a$.
Then, $\parallel Vx\parallel = \parallel A^*a+Vv\parallel$
To make $V$ as unitary $\parallel Vv\parallel$ should be $\parallel v\parallel$
Also, $Vv$ must orthogonal with $Range(A^*)$
But I can't how to define this $Vv$ so that we can make sure existence of mapping $V$ which is unitary.
One has $\ker(R) = \ker(A^*)$, hence $ran(R) = ran(R^*) = \ker(R)^\perp = \ker(A^*)^\perp$. It follows that the space $E$ is the orthogonal direct sum \begin{equation} E = \ker(A^*)^\perp \perp \ker(A^*)=ran(R)\perp \ker(A^*) \end{equation} You already defined $V$ on $ran(R)$, it remains to define it on $\ker(A^*)$. As $V$ sends $ran(R)$ onto $ran(A^*)=\ker(A)^\perp$ and you want a unitary operator, you need to send $\ker(A^*)$ to $\ker(A)$. Take any linear isometry \begin{equation} \varphi: \ker(A^*)\to\ker(A) \end{equation} and define $V x = \varphi(x)$ for all $x\in\ker(A^*)$. This extension of $V$ should solve the problem.