Unitary Transformation of Uniform random vector Is the following true:

For any unitary matrix $U \in \mathbb{R}^{n \times n}$, there exists
  a distribution on a random vector $X\in \mathbb{R}^{n }$ such that 
  \begin{align} Y= UX \end{align} where $Y_i$ ($i$-th component of $Y$)
  are independent and  unform on $[a_i,b_i]$.

What  I did.  Let $S$ be a cartesian product of $[a_i,b_i]$. In other words, $S$ is the support of the random vector $Y$. Therefore,
\begin{align}
f_Y(y)=\frac{1}{Vol(S)}, y \in S
\end{align}
By the linear transformation of we have that
\begin{align}
f_X(x)=\frac{1}{det(U)Vol(S)}=\frac{1}{Vol(S)}, x \in S'
\end{align}
where the last step follow since $det(U)=1$ and where $S'=\{x: U^{-1}y \in y\in S\}$.
My Questions:  Does this Constitute the proof? Can we say something about the marginal of $X_i$ ($i$-th component of $X$). 
Thanks.
 A: The only incorrect place is that $det(U)$ does not necessarily equal to $1$. It can be $\pm 1$, so you need $|det(U)|$. 
I repeat the main part of your solution. Let 
$$f_Y(y)=\frac{1}{Vol(S)}, \, y \in S,$$
and $X=U^{-1}Y=U'Y$, where $U'=U^{-1}$ is a transpose matrix of $U$, then
$$f_X(x)=\dfrac{1}{|det(U')|}f_Y\left((U')^{-1}x\right) = \dfrac{1}{|det(U)|}f_Y(Ux)=\frac{1}{Vol(S)},\, x\in S'.
$$
The area $S'=\{x: Ux\in S\}=\{x=U'y: y\in S\}=U'S$.
The transformation $S\mapsto U'S$  is a unitary transformation (rotation and/or reflection) in $\mathbb R^n$. If we rotate/reflect a rectangle $[a_1, b_1]\times\ldots\times[a_n,b_n]$, we some cases can obtain the new cartesian product as a result (under some rotation angles or after reflection). And this cases the independence of coordinates is preserved. But in general the result of this transform leads to dependent r.v.
If you want to see marginals, I doubt that is can be written in closed form. Say, 
$$X_i=(U'Y)_i=\sum_{j=1}^n u'_{i,j}Y_j=\sum_{j=1}^n u_{j,i}Y_j$$
This is a sum of independent r.v. with different Uniform distributions. The fact that $\sum_{j=1}^n u_{j,i}^2=1$ cannot simplify this task since $a_j, b_j$ are arbitrary.
You can see the discussion on this subject here and some results here. The second paper find the convolution of Uniform distributed p.d.f. on $[0,a_i]$. Simple shift provides a general case.
