We defined conformal and isometric maps for surfaces $f,g: \Omega \subset \mathbb{R}^2 \rightarrow S \subset \mathbb{R}^3$.
Under a reparametrization of $f$ I understand a diffeomorphism $\Phi : M \subset \mathbb{R}^2 \rightarrow \Omega$ such that our surface is given by $f \circ \Phi$.
So $f$ is conformal to $g$ if there is a positive function $\lambda$ such that $Df^T Df = \lambda Dg^T Dg$ and isometric if $\lambda =1$.
Now, I was wondering how I can find out if there is a reparametrization of $f$ and $g$ sucht that they are locally/globally conformal/isometric?
We had one theorem heading in this direction that was like: If the two surfaces that have the same Gauß curvature locally, then they are locally isometric.
But what about the global case and what about conformal reparametrizations?
As I said in my comment, isometric or conformal type of a surface in $R^3$ is independent of a parameterization.
This is an answer in the conformal setting (isometric setting is hopeless). You need:
Uniformization Theorem: Every simply-connected Riemann surface is conformal either to the unit disk, or to the complex plane or to the 2-sphere.
In view of this theorem, if you have two simply-connected surfaces $S_1, S_2$ in $R^3$, equipped with the induced Riemannian metric, then they have one of the above three conformal types. To show that they are conformal to each other, you need a bit of luck. For instance, if each surface appears as a graph of a smooth bounded function over a bounded ( simply-connected) domain, then one can prove that they are conformal to the unit disk and, hence, to each other.
You can read more about uniformization and the 3 types of simply-connected Riemann surfaces (and how to tell them apart!) in the book by Ahlfors and Sario "Riemann surfaces".