Is it possible to convert a Rayleigh distribution into a Gaussian distribution? ...and how might we do this? If possible, I am curious if outliers in the Rayleigh distributed data would also remain outliers in the new Gaussian distributed data. Thanks.
 A: If $R$ is a Rayleigh random variable and $\Theta \sim U[0,2\pi)$ is independent of $R$, then $X=R\cos \Theta$ and $Y=R \sin \Theta$ are independent
zero-mean normal random variables with identical variance 
$\sigma^2 = \frac{1}{2}E[R^2]$.  Thus, if you transform your data set as
$$\{r_1, r_2, \ldots, r_n\} \longrightarrow \{r_1\cos \theta_1, 
r_2\cos \theta_2, \ldots r_n\cos \theta_n\}$$
(similarly for $Y$ but using $\sin \theta_i$) where 
$\{\theta_i\colon 1 \leq i \leq n\}$ is a 
data set that you create as a sequence of independent samples drawn from
$U[0,2\pi)$,, then the resulting data set is exactly a collection of $n$
samples from a $N(0,\sigma^2)$ distribution.  Note that in contrast to
the monotone transformations suggested in @Glen_b's answer, the resulting data
set has both positive and negative numbers in it.


*

*A Rayleigh-distributed data set can have outliers that are very
large, but since these will get multiplied by a cosine, which has
magnitude less than $1$, they might not be outliers any more
unless the cosine has magnitude close to $1$, as noted in Glen_b's
comment.  Note also that if $\cos \theta_i$ is close to
$-1$, then this outlier will still be an outlier but in the other
tail.  

*If $E]R^2]$ is quite large, then a Rayleigh-distributed
data set can also have very small outliers:  sample values
that are very close to $0$ while almost all of the other sample
values are closer to the (large) sample mean.  Such outliers will
become very close to the mean when they undergo the transformation
that I suggest above. With Glen_b's transformation, such outliers
will remain small outliers.
In short, the transformed data set does not enjoy the property that outliers 
in $\{r_i\colon 1 \le i \leq n\}$ are outliers in the transformed data set $\{r_i\cos \theta_i\colon 1 \le i \leq n\}$.  If you are willing to have
a transformed data set that is twice as large, then the set
$$\{r_1\cos \theta_1, 
r_2\cos \theta_2, \ldots r_n\cos \theta_n\} \cup 
\{r_1\sin \theta_1, 
r_2\sin \theta_2, \ldots r_n\sin \theta_n\}$$
is a set of $2n$ independent samples drawn from an $N(0,\sigma^2)$
distribution, and since 
$\max \{|\cos \theta |, |\sin \theta |\} \geq 1/\sqrt{2}$
you are guaranteed that each large outlier gives you two numbers
at least one of which might well still be an outlier, since outlying
is, like beauty, in the eye of the beholder. Is something just
$20\%$ larger than the next smaller value an outlier? or would you
insist on $50\%$ larger? I suppose it depends on the scale factors
etc.
A: If you know the Rayleigh parameter, then the conversion to a standard normal is readily achieved by the probability integral transform followed by an inverse normal cdf. If $X\sim\text{Rayleigh}(\sigma)$, with cdf $F_\sigma(x)$, then $F_\sigma (X)$ is uniform, and $\Phi^{-1}(F_\sigma (X))$ is standard normal (where $\Phi$ is the standard normal cdf).
If $\sigma$ is unknown we are left with one kind of approximation or another (even estimating $\sigma$ involves approximation). Since the square of a Rayleigh random variable is a special case of the gamma, the Wilson-Hilferty transformation (cube root in the case of the gamma) should produce a good approximation of normality. That is, if $X\sim \text{Rayleigh}$, then $X^{2/3}$ should be fairly normal looking. 

In practice it looks like a slightly smaller power, somewhere near 0.6, might be a little closer.
$\text{ }$
Here's a comparison of the exact transformation (x-axis) and the three power transformations above (y-axis):

The black is the 0.6 power, the red is the 2/3 power and the green is the 1/2 power. The one that most closely reproduces the exact transformation should lie closest to a straight line ... and that looks to be the green line.  
(Added in edit: I've checked more carefully; of those three, the green line is nearest to straight overall, but the black line is straighter in the right tail. All three give distributions that are pretty nearly normal - but I should have asked why you needed normality.)
--
An additional discussion on powers of exponential - and hence of Rayleigh - variables: 
Powers of exponential random variables are distributed as Weibull. Specifically, if $X$ is exponential, $X^\frac{1}{k}$ is Weibull with shape parameter $k$. 
So cube roots and fourth roots of exponentials (Rayleigh variables to the powers $\frac{2}{3}$ and $\frac12$ respectively) will be Weibull with shape parameters $3$ and $4$.
While no Weibull is symmetric, particular choices for $k$ will produce Weibull distributions with zero skewness (different choices for different measures of skewness).
Here are the values of $k$ that have $0$ skewness for several skewness measures and the corresponding power ($p$) of the Rayleigh:


*

*3rd-moment skewness:   $\:\,\, k=3.60,\, p=0.56$  

*mean-median skewness:  $k=3.44,\, p=0.58\,$ (second Pearson skewness)

*mean-mode skewness:    $\:\,\, k=3.31,\, p=0.60\,$ (first Pearson skewness)

*mode-median skewness:  $k=3.26,\, p=0.61$ 


So it's little surprise that a value of $p$ near $0.6$ would appear suitable.
--
In respect of the issue with outliers:
Can you define what you mean by 'outlier'? If the data has values that don't actually come from the Rayleigh distribution, the question of transforming Rayleigh-distributed data is irrelevant, since you don't have Rayleigh-distributed data.
So if you have values from some distribution that's not Rayleigh (with values that a discrepant for a Rayleigh model) and you transform to normality as if it were Rayleigh, then you definitely end up with a non-normal result ... one which will likely have discrepant-looking values relatively to a normal model.
In the absence of a specification of what makes an outlier, here's an example:

This is the same data as above (a large sample from a Rayleigh distribution with $\sigma=1$). In the second panel I have added four outliers (at 6, 10, 30 and 100), marked in red. In the third panel, we see the effect of the power $p=\frac{2}{3}$ and in the last panel the effect of the power $p=\frac{1}{2}$. Note that for the Rayleigh data, $\mu+3\sigma$ is at about 3.22. The value at 6 is already far enough away as to be considered highly unlikely even in a quite large sample with that level of skewness -- an outlier in some sense. You can see that the transformation brings it back toward the main part of the data, so that by the last panel it's visually a somewhat mild outlier, fairly borderline given the sample size (about 4.5 sd's above the mean).
The value 10 (on the original scale), while clearly an outlier, is noticeably less discrepant but still clearly inconsistent with the idea that the data are normal, and the larger values even more so. 
So they do in some sense "come in" -- but as to whether the status of them being 'outliers' has changed depends very much on how you define 'outlier'.
--
The suggestion of a fourth root for a gamma* with small values of the shape parameter is in Hawkins and Wixley (ref below). While the Wilson-Hilferty works best over a wide range of gamma distributions, the Hawkins-Wixley does seem to do a little better in some parts of the low-end (smaller values of the shape parameters). That fourth root of a gamma corresponds to a square root of a Rayleigh.
*(NB chi-square is a gamma with a particular scale, and when looking at power transformation, the scale won't matter. So even though both references seem to be about chi-square distributions, their conclusions apply to gamma distributions more generally) 
Wilson, E. B., and Hilferty, M. M. (1931),
"The Distribution of Chi-Squares,"
Proceedings of the National Academy of Sciences, 17, 684–688.
Hawkins, D. M., and Wixley, R. A. J. (1986),
"A Note on the Transformation of Chi-Squared Variables to Normality,"
The American Statistician, 40, 296–298.
