Independence under linear transformations What is the (largest) set of matrices $\mathcal C\subset \mathbb R^{m\times n}$ ($m\le n$) for which the following statement is true?

Let $x_1,\ldots,x_n\in\mathbb R$ be independent random variables. Then also $z_1, \ldots, z_m$ are independent, given $z=Ax$ with $A\in\mathcal C$ 

It obviously holds when $A$ is a (rectangular) diagonal matrix. Is it also true if the rows are (1) orthogonal (2) linearly independent?
Conjecture: It is only true for matrices for which it is trivially true, i.e. when there is no interaction between any of the $x_i$:
$$\mathcal C = \{DP\mid D\in\mathbb R^{m\times n} \text{ diagonal, } P\in\mathbb R^{n\times n} \text{ permutation matrix} \}$$
 A: This is the set $C_{m,n}$ of matrices that have at most one nonzero entry in each column.
Let us rephrase the issue in the question as a property of an $m\times n$ matrix $A:$

$\mathcal{P}(A):$ for all independent random variables $x=(x_1,\ldots,x_n)$ defined on a given space $\Omega,$ $z=Ax$ is independent.

My assertion requires two demonstrations: namely, that $A\in C_{m,n}$ implies $\mathcal{P}(A)$ and then that $A\notin C_{m,n}$ implies $\mathcal{P}(A)$ is false.
First, when $A\in C_{m,n}$ and the $x_j$ are independent, we have to show the $z_i$ are independent.  This can be done with an induction on $m.$  When $m=1,$ $\{z_1\}$ is trivially independent. Now assume the result for $m-1.$ The structure of $A$ is such that $z_m$ can be expressed in terms of the $x_j$ for which $A_{mj}\ne 0$ and all the other $z_i,$ $i\lt m,$ can be expressed in terms of the remaining $x_j.$ Because the $x_j$ are independent, $z_m$ is therefore independent of the remaining $z_i,$ whence $\{z_1,\ldots,z_m\}$ is independent.  That establishes the first claim.
Second, suppose $A \notin C_{m,n}.$  This means there is a column $j$ and two rows $i,i^\prime$ for which $A_{ij} \ne 0$ and $A_{i^\prime j}\ne 0.$ We may therefore write
$$z_i = A_{ij} x_j + y_i,\quad z_{i^\prime} = A_{i^\prime j} x_j + y_{i^\prime}$$
for variables $y_i$ and $y_{i^\prime}$ formed out of the remaining $x_{j^\prime},$ $j^\prime\ne j.$  Setting all such $x_{j^\prime}$ to be constant and selecting $x_j$ to have unit variance, compute
$$\operatorname{Cov}(z_i,z_{i^\prime}) = A_{ij}A_{i^\prime j} \ne 0,$$
showing $z_i$ and $z_{i^\prime}$ are not independent, which means $\mathcal{P}(A)$ is false, QED.
A: For orthogonal columns, in general it's not. Let $$A=\begin{bmatrix}1 &-1\\1&1\end{bmatrix}$$
This results in $z_1=x_1-x_2,z_2=x_1+x_2$. These two aren't independent in general. For example, assume $x_i$ are Bernoulli RVs, then if $z_2=2$, $z_1$ is definitely $0$, which means there is dependency.
For the linear independence, see @David's counter-example. 
