Edit: No, it does not hold in general. For some reason I always tended to use that $V$ commuted with $X^*$, which in general doesn't have to be true. A counterexample:
Let $\mathcal{H}=\ell^2(\mathbb{N})$. Let $V$ be the right shift on $\mathcal{H}$, i.e. $V\delta_n=\delta_{n+1}$ where $\delta_n(m)=\left\{\begin{matrix} 1 & \mbox{if }n=m\\0 & \mbox{if }n \neq m\end{matrix}\right.$. Let $X=2+V$ (so $X\delta_n=2\delta_n+\delta_{n+1}$) then $XV=VX$. However, you can check that $X^*V \neq VX^*$. Now $X$ is invertible, so $U$ is unitary. But then $UV=VU \Leftrightarrow X^*V=VX^*$, which is false.
So the following only holds if $V$ commutes with $X^*$:
Let $\mathcal{X}$ be the von Neumann algebra generated by $X$. One can show that in this case $U\in \mathcal{X}$. As clearly $V$ is an element of the commutant of $\mathcal{X}$ and $U \in \mathcal{X}$, we find that $VU=UV$. So I don't think you need the fact that $V$ is an isometry...
I'm assuming you are asking about the uniqueness of $U$ (at least that's what Rudin's exercise asks).
The result is that $U$ is unique if and only if $T$ has dense range.
Note that, since $U$ is a unitary, $$\tag{1}(\mbox{Im}\,T^*)^\perp=\ker T=\ker T^*=(\mbox{Im}\,T)^\perp.$$
Assume first that $U$ is unique. Let $P$ be the orthogonal projection onto $(\mbox{Im}\,T)^\perp$. Note that $U$ maps $\overline{\mbox{Im}\,T}$ into $\overline{\mbox{Im}\,T^*}=\overline{\mbox{Im}\,T}$. This means that $U(I-P)=(I-P)U(I-P)$. Multiplying by $P$ on the left, $PU(I-P)=0$. We can repeat the same reasoning for $T^*$, to get $PU^*(I-P)=0$. So $PU=PUP$, $PU^*=PU^*P$, and taking adjoint we get $UP=PUP=PU$. Now consider, for $\lambda\in\mathbb C$ with $|\lambda|=1$, $V=\lambda P+U(I-P)$; this is a unitary, since
$$
V^*V=(\overline\lambda P+(I-P)U^*)(\lambda P+U(I-P))=I+2\mbox{Re}\,\overline\lambda PU(I-P)=I,
$$
and
$$
VV^*=(\lambda P+U(I-P))(\overline\lambda P+(I-P)U^*)=P+U(I-P)U^*=P+(I-P)=I.
$$
Also, $VT=\lambda PT+U(I-P)T=0+(I-P)UT=(I-P)T^*=T^*$. By the uniqueness $U=V$, that is $\lambda P+U(I-P)=U=UP+U(I-P)$. So $\lambda P=UP$. As $\lambda$ was arbitrary, this shows that $P=0$. That is, $T$ has dense range.
Conversely, if $T$ has dense range, then $T$ and $T^*$ are injective, by $(1)$. Given $y\in\mbox{Im}\,T$, there exists $x$ such that $y=Tx$. This $x$ is unique by the injectivity of $T$. Then $Uy=UTx=T^*x$, so $U$ is defined uniquely on a dense subspace, and so it is defined uniquely.
Best Answer
Since $T^2=I$, you have $T^{-1}=T$. From $T=U|T|$, we get $U=T|T|^{-1}$. Then $$ |T|^{-1}=(T^*T)^{-1/2}=[(T^*T)^{-1}]^{1/2}=[T^{-1}{T^{-1}}^*]^{1/2}=(TT^*)^{1/2}. $$ Now the key fact is that $T(T^*T)^n=(TT^*)^nT$, and as the square root is a limit of polynomials with no constant term, $$ T(T^*T)^{1/2}=(TT^*)^{1/2}T. $$ Thus $$ |T|^{-1}T=(TT^*)^{1/2}T=T(T^*T)^{1/2}=T|T|. $$ Thus $$ U^2=T|T|^{-1}T|T|^{-1}=TT|T||T|^{-1}=I. $$