Note: this question is inspired by this one: Why $\arctan x$ not equal to $\arcsin(x)/\arccos(x)$?
In the linked question, it is said that if $f=\frac{g}{h}$ and $f$, $g$ and $h$ are invertible, then $f^{-1}\neq \frac{g^{-1}}{h^{-1}}$ in general. This is clear that it is almost always the case after considering a few examples, but it may be possible that there exists invertible functions $f$, $g$ and $h$ such that $f=\frac{g}{h}$ and $f^{-1}=\frac{g^{-1}}{h^{-1}}$.
I tried a few functions (linear functions, exponentials,...) but without an example.
Are there any functions satisfying the above property?
Mastrem's comment shows that a trivial solution exists with $f=g=h=id:\{1\}\to\{1\}$. I am looking for a non-trivial example. Functions can have any domain and codomain, including finite ones, as soon as they have more than one point.
One way to create examples of this phenomenon is to find an invertible function $f$ such that $g(x)=f(x)f^{-1}(x)$ is its own inverse, and then take $h(x) = f^{-1}(x)$. For example, here is one construction on the six-point domain $\{\frac13,\frac12,\frac23,\frac32,2,3\}$ where $g(x)=x$: \begin{align*} f(\tfrac13) &= \tfrac23 & g(\tfrac13) &= \tfrac13 & h(\tfrac13) &= \tfrac12 \\ f(\tfrac12) &= \tfrac13 & g(\tfrac12) &= \tfrac12 & h(\tfrac12) &= \tfrac32 \\ f(\tfrac23) &= 2 & g(\tfrac23) &= \tfrac23 & h(\tfrac23) &= \tfrac13 \\ f(\tfrac32) &= \tfrac12 & g(\tfrac32) &= \tfrac32 & h(\tfrac32) &= 3\\ f(2) &= 3 & g(2) &= 2 & h(2) &= \tfrac23 \\ f(3) &= \tfrac32 & g(3) &= 3 & h(3) &= 2 \end{align*}