I am looking for an example of the following:
Find a bijective, differentiable function $f$ and continuous probability density functions $q\_1\ne q\_2$ such that $f\_\*q\_1=p=f\_\*q\_2$, where $f\_\*$ is the pushforward density and $p$ is continuous as well. What if continuity is strengthened to differentiability?


Edit: Intuitively this seems impossible, just by continuity considerations; e.g. pick a neighborhood where $q\_1$ and $q\_2$ differ, and invoke bijectivity of $f$.



Transforming two smooth densities to the same density
This is impossible if $f$ is injective, without further assumptions such as bijective, differentiable, etc. Let $Q\_1,Q\_2$ be probability measures on a measurable space $(\Omega, \mathcal{F})$, and assume $f\_\* Q\_1 = f\_\* Q\_2$ for some injective (bimeasurable) $f : (\Omega,\mathcal{F}) \to (\Xi,\mathcal{G})$. For any $A\in \mathcal{F}$, definitions give


$$
Q\_1(A) = f\_\* Q\_1 (f(A)) = f\_\* Q\_2 (f(A)) = Q\_2(A).
$$


Thus, $Q\_1 = Q\_2$.


