Stochastic inverse Let $X_t$ be a semi-martingale and $H_t$ be a predictable process and $g$ be a measurable bijective function with measurable inverse.  Does there exist a function $f(h,x)$ satisfying
$$
\int_0^Tf(H_t,X_t) dg(X_t) = \int_0^TH_t dX_t?
$$
If not what conditions do we need for that to hold?
 A: The LHS might not make sense if g is very irregular.
Consider, pth-Holder function $g(x)=x^{p}$ for $0<p<1$ and $\frac{1}{p}$ odd integer and Brownian motion $X_{t}=B_{t}$, then the process $g(X_{t})= (B_{t})^{p}$ for $p<\frac{1}{3}$ is not even in the usual regime for rough integrals (Uniqueness of solutions of Young differential equations). So we would also need to formulate the LHS in some way.
Generally, we also need $C^2$/Convex in order to apply change of variables see here https://almostsuremath.com/2020/10/12/the-ito-tanaka-meyer-formula/
In even weaker settings of rough paths, there is still an Ito formula (In Proposition 6.9 "Rough Path Theory" by Andrew L. Allan).

Even for just Brownian motion for the Ito-formulation, in most cases of $g$ there will be a drift-term (if there was a formula with no-drift term, by taking expectation or using Ito-isometry we would likely get contradictions to the Ito formula).
But if you interpet the RHS as the Stratonovich-formulation there is the regular chain rule
$$  g'(W_t) \circ \mathrm{d} W_t = dg(W_t),$$
and so one can use the above Ito-formulas to get something close to the OP statement (see here for semimartingales and Stratonovich).
