If $f(x)=g(x)f(h(x))$ with $g$ and $h$ known, can I obtain $f$?

EDIT-giving context: I was trying to find an explicit expression for a probability distribution using the Chapman-Kolmogorov equation. After using a saddle point approximation I reached an expression of the type $f(x)=g(x)f(h(x))$ being $f(x)$ the probability distribution in which I am interested and g(x) and h(x) two known non-linear functions (and well behaved). Since I am not used to work with this kind of recurrent equation, I was curious about a rather general question.

Say that there exists an unknown function $f$ that fulfills $\begin{equation} f(x)=g(x)f(h(x) \end{equation}$ and $\int dx f(x)=1$ with $g$ and $h$ being two known non-linear and well-behaved functions. Can the above equation be used to determine $f$? How?

The equation is linear in $f$, so the answer can never be unique unless it's identically $0$ (which is always a solution).

An example where the only solution is $0$ is $h(x) = x$, $g(x) = 2$.

In general, $\mathbb R$ is partitioned into orbits under the map $h$: $x$ and $y$ are in the same orbit if $h^n(x) = h^m(y)$ for some positive integers $n$ and $m$ (where $h^n$ means $h$ iterated $n$ times). On each orbit, $f$ is determined up to a multiplicative constant. On a periodic (or eventually periodic) orbit, $f$ may be forced to be $0$. There are no constraints linking different orbits.