Suppose $m$ is a smooth Riemannian metric on $\mathbb{S}^2$, the uniformization theorem of surfaces tell us that $m$ is conformally equivalent to the standard round metric. Formally this says that there exists a diffeomorphism $f$ of $\mathbb{S}^2$ and a function $h: \mathbb{S}^2 \to \mathbb{R}$ such that $f^{*}(m) = e^{h}m_0$, where $m_0$ is the standard round metric and $f^{*}$ denotes pullback by $f$.

Observe that $f$ and $h$ are not unique. One can compose $f$ by Mobieus transformations and obtain different $f$ and $h$ that way. Nonetheless, one would expect that it is possible to find $f$ and $h$ that just depend on how my Riemannian metric $m$ is.

Formally, my question is: Given a smooth metric $m$ on $\mathbb{S}^2$. Can one obtain a pair $(f,h)$ uniformizing $m$ whose $C^r$ norm (measuring up to $r$ derivatives) depends (say polynomial, exponential, etc) on the $C^r$ norm of the metric $m$ (as a 2 tensor)?

Any result vaguely related to this question will be greatly appreciated.