"Convert" Rayleigh random variable into a Uniform random variable? I have a nested question of sorts. My first question, is that I am wondering if it is possible to 'convert' a Rayleigh random variable into a uniform random variable, and how one may do this. 
Strongly related and dependent on this question though, is: if such a conversion was possible, would 'corrupt data', being defined here as outliers in the original Rayleigh PDF, also remain 'corrupt' outliers in the uniform distribution?
 A: 
Your assertion that "a $\chi^{2}$ random variable (with 2 degrees of freedom; aka Rayleigh random variable)"
  is incorrect: a  $\chi^{2}$ random variable with two degrees of freedom is
  an exponential random variable with mean $2$, and not a Rayleigh random
  variable.

Following up user603's comments and my own comments about distributions,


*

*If $X$ is a $\chi^2$ random variable with two degrees of freedom, then
its CDF is $F_X(x) = [1 - \exp(-x/2)]\mathbf 1_{x\in [0,\infty)}$ and so
if $x_1, x_2, \ldots, x_n$ are samples of $X$, then $y_1, y_2, \ldots, y_n$ are 
samples of 
a $U[0,1)$ random variable where $y_i = 1-e^{-x_i/2}$, $1 \leq i \leq n$.

*If $Z$ is a Rayleigh random variable, its CDF is
$F_Z(z) = [1 - \exp(-z^2/2)]\mathbf 1_{z\in [0,\infty)}$
and so if $z_1, z_2, \ldots, z_n$ are samples of $Z$, then 
$w_1, w_2, \ldots, w_n$ are 
samples of 
a $U[0,1)$ random variable where $w_i = 1-e^{-z_i^2/2}$, $1 \leq i \leq n$.

*$\sqrt{X}$ is a Rayleigh random variable, and so if we have samples
$x_1, x_2, \ldots, x_n$  of $X$, then $\sqrt{x_1}, \sqrt{x_2}, \ldots, \sqrt{x_n}$
are samples of a Rayleigh random variable, and regardless of whether we choose to
map $x_i \mapsto 1-e^{-x_i/2}$
or to map $\sqrt{x_i} \mapsto 1-e^{-\left(\sqrt{x_i}\right)^2/2}$, we get the
same $y_1, y_2, \ldots, y_n$ as samples of a $U[0,1)$ random variable.
With regard to your question about outliers, note that the mean of $X$ is $2$
which maps to $1-e^{-1} \approx 0.632$ and so outliers, say any sample larger
than $10$, are mapped into a narrow range $[0.99326\ldots, 1)$. Remember that
sample values close to $0$ are not outliers, since the density of $X$  is
a monotone decreasing function on $\mathbb R^+$ and so values close to $0$ 
are highly likely and not rare at all.
