For simplicity let $X$ be a uniformly distributed random variable on $[0,1]$. Does there exist a pair of measurable functions $f$ and $g$ such that $f(X)$ and $g(X)$ are independent random variables?
Building off of Parcly's answer, if $x\in [0,1)$ is expressed in base 2, $x=0.b_1b_2\dots_2$, then $f(x) = 0.b_1b_3b_5\dots$ and $g(x) = 0.b_2b_4b_6\dots$ are both limits of independant random variables so they at least measurable. In particular, $f_n = 0.b_1b_3...b_{2n+1}\uparrow f$ and similarly $g_n\uparrow g$. Thus for every $y$ and $z$, \begin{equation} P(f<y\textrm{ and } g<z)=\lim P(f_n<y\textrm{ and }g_n<z)=\lim P(f_n<y)P(g_n<z)=(\lim P(f_n<y))(\lim P(g_n<z))=P(f<y)P(g<z) \end{equation} where the first and last equalities are due to the regularity of the measure. Additionally, each $f$ and $g$ are bijections of the interval with itself and so for any other measurable functions $h(x)$ and $j(x)$, we have $h\circ f$ and $j\circ g$ independent random variables. What's more, $h\circ f$ and $h\circ g$ are independent while it is not true that $h(x)$ and $j(x)$ are independent when mappings of $X$ alone.
For a given variate $x=0.b_1b_2b_3b_4\dots_2$ from $X$, let $f(x)=b_1$ and $g(x)=b_2$. Since each bit in $x$ is independent of each other, it follows that $f(X)$ and $g(X)$ are two independent random variables with an equal probability of taking up the values 0 and 1.