Can any finite variance distribution be transformed into the normal distribution?
For example, we can translate the uniform and exponential distributions into the normal distribution.
EDIT: Let's specify that the functions to be transformed are differentiable over their entire domain.
On
@kimchi lover wants my comment to be posted as an answer: so here goes:
NO. You cannot transform an arbitrary distribution into a Normal Distribution by a continuous transform of domain ($x$ values).
A simple proof: Some distributions have different numbers of local maxima (and minima). No continuous transformation will change the number of such local maxima and minima.
Thus any of an infinite number of distributions that have two local maxima cannot be transformed into a Gaussian (which has just one maximum).
And likewise for distributions with more local maxima.
"No" to the question as originally phrased, without the clause "the functions to be transformed are differentiable over their entire domain". Continuous distributions can be so transformed, but not discontinuous ones. That is, for a random variable $X$, if there exists any number $a$ such that $P(X=a)>0$ then there is no transformation $g$ such that the random variable $g(X)$ is normally distributed. Finite variance or not. Your examples (exponential, uniform) are continuous, but other well-known distributions (binomial, geometric, Poisson) are not.
The edited version of the question is imprecise. I assume the functions that must be differentiable over their domains are the cumulative distribution functions. This has the effect of ruling out the OP's two examples (uniform and exponential) whose distribution functions are not differentiable at $0$, but does lead to an easy answer: yes.
The important concept here is continuous random variable, which means one whose cumulative distribution function (or "cdf") is continuous. It is a theorem that if $F_X$ is a continuous cdf and $F_Y$ is any cdf, one can generate a random variable $Y$ with cdf $F_Y$ by transforming a random variable $X$ with cdf $F_X$, by taking $Y=g(X)$, where $g(t)=F_Y^{-1}(F_X(t))$ is a monotone function. (See this and that for wikipedia evidence.) If, as the OP wants, $F_X$ is assumed differentiable, and the target $F_Y$ is the standard normal cdf $\Phi$, then $g$ will be differentiable, too.
The condition that $X$ is continuous is not the same as its probability density function being continuous. It is the same as requiring that $P(X=a)=0$ for all real $a$.
Another answer to this problem asserts that if the density function of $X$ has one hump, so must the density of $g(X)$ for any continuous $g$. This is easily seen to be false. If $X$ has the Cauchy distribution, with density $1/(\pi(1+x^2))$, and $g(x)=x+a\sin(x)$, for $a=9/10$, the density of $Y=g(X)$ will have many humps, near $x$ values that are multiples of $\pi$. This is because this particular $g$ has the effect of periodically puckering up and stretching the $x$-axis, which shows up as ripples in the graph of the derivative of $g$. The density functions for $X$ and $Y$ are related as $f_Y(y)=f_X(x)/g'(x)$ with $y=g(x)$; when $g'(x)$ is small $f_Y(y)$ gets a hump. (It is easy to simulate this on a computer and inspect the resulting histograms: they are quite striking.)