Assume real valued random variables $X,Y,Z$ such that $Z = f(X)$ for an injective, smooth $f$ with ${f^{-1}}'\neq 0$.
Suppose we have a joint density $$p(y,x) = p(y|x) p_x(x) = g(y,f(x))p(x)$$ for some $g$ (for example $g = \mathcal{N}(y,g(x),\sigma)$).
We then also have a joint density
$$p(y,z)=p(y|z)p_z(z) = g(y,z)p_x(f^{-1}(z)) |{f^{-1}}'(z)|$$
Now intuitively, the maxima of these joint densities should coincide, i.e. if for fixed $y$, $x^*$ is a maximum of $p(y,x)$ then also $z^*=f(x^*)$ should be a maximum of $p(y,z)$. But it seems not to be correct:
If we apply logarithms to both densities and derivate, we obtain
$$\frac{d}{dx} \ln p(y,x) = (\frac{d}{du}\ln g(y,f(x)))f'(x) + \frac{d}{dx} \ln p(x)$$ $$ \frac{d}{dx} \ln p(y,z) = (\frac{d}{du}\ln g(y,z)) + (\frac{d}{dx}\ln p_x(f^{-1}(z))) ({f^{-1}}'(z)) + \frac{d}{dz} \ln |{f^{-1}}'(z)| ,$$ where by $du$ I denoted derivating w.r.t. the second component for clarity.
Now, I don't really see how to proceed to show equality. For example, the second equation is zero iff $$(\frac{d}{du}\ln g(y,z))\frac{1}{({f^{-1}}'(z))} + (\frac{d}{dx}\ln p_x(f^{-1}(z))) + \frac{\frac{d}{dz} \ln |{f^{-1}}'(z)|}{{f^{-1}}'(z))}=0$$
so now plugging in $z^* = f(x*)$ would yield
$$ 0 + 0 + \frac{\frac{d}{dz} \ln |{f^{-1}}'(z^*)|}{{f^{-1}}'(z^*))},$$
where the last term does not seem to be $0$ necessarily.