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This is impossible if $f$ is invertible, 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 invertible (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, $Q_1 = Q_2$.

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, $Q_1 = Q_2$.

This is impossible if $f$ is invertible, 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 invertible (measurable) $f : \Omega \to \Xi$. 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, $Q_1 = Q_2$.

This is impossible if $f$ is invertible, 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 invertible (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, $Q_1 = Q_2$.

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, $Q_1 = Q_2$.

This is impossible if $f$ is invertible, 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 invertible (measurable) $f : \Omega \to \Xi$. 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, $Q_1 = Q_2$.