Möbius–Transformed Hamilton–Jacobi Theory

A typical question might be:

Can you find a system where the HJ-equation becomes simpler under a Mobius conformal symplectic structure?

I don't know. Let's sketch something for the simplest case. Take a dimensionless HO with $H(q,p) = \frac{1}{2}(p^2 + q^2)$ and in complex form $z = q + ip$ with $H = \frac{1}{2} \mid z \mid ^2$. Take a Mobius map $f = \frac{az+b}{cz+d}$ with $ad-bc \neq 0$. The canonical form is $\omega = dq \wedge dp = \frac{i}{2} dz \wedge d \overline{z}$.

Let's rewrite H in terms of w which we obtain by transforming z. $z = (dw-b) / (-cw +a)$ st $|z|^2 = \mid dw-b |^2 / \mid a-cw \mid ^2$ where z becomes a function of w. $H(w,\overline{w}) = \frac{1}{2} \mid dw -b \mid^2 / \mid a -cw \mid ^2$. Short: $H(q,p) \to H(z, \overline{z}) \to H(w, \overline{w})$. The purpose of this is to map the circular orbit into a straight line, with $H(s) = \frac{1}{2} (ds-b)^2 / (a-cs)^2$. (orbit becomes a line, jettison the complex number, s is just parametrization of the line.) Identify $w=s$ on the physical trajectory. On this orbit the Hamiltonian must equal the energy: $\frac{1}{2} (ds-b)^2 / (a-cs)^2 = E$ with $\partial S / \partial s = \pm \sqrt{ 2E } (a-cs) / (ds -b)$.

Integrating $S(s) = \pm \sqrt{ 2E } \int (a-cs) / (ds -b) \, ds = \pm \sqrt{ 2E } [(-c/d)s + (ad-bc) /d^2 * \ln|ds-b|]$.

So isn't this a simplification after all? I'll keep adding more examples when I have the time.