This is impossible if $f$ is invertible. Let $Q_1,Q_2$ be distributions, and assume $f_* Q_1 = f_* Q_2$ for some invertible (measurable) $f$. For any Borel set $B$, definitions give
$$ Q_1(B) = f_* Q_1 (f(B)) = f_* Q_2 (f(B)) = Q_2(B). $$
Thus, $Q_1 = Q_2$.