Let $g$ be any Riemannian metric on the 2-sphere $S^2$ and let $g_0$ be the round metric (of constant curvature $K$, say).
Is it true that there exists a smooth positive function $\lambda:S^2\to \mathbb{R}$ such that $g_0=\lambda^2g$?
The version of the uniformization theorem that I am familiar with implies that there exists a diffeomorphism $f:S^2\to S^2$ and a smooth function $u:S^2\to \mathbb{R}$ such that the pullback metric $f^{\ast}g$ is conformal to the round metric, i.e. such that $g_0=u^2(f^{\ast}g)$.
So my question is equivalent to the following question: can the diffeomorphism $f$ be taken to be the identity?
The reason I think this might be true is that a 2-dimensional manifold has constant curvature iff it has constant scalar curvature. For $n\geq 3$, any metric on a closed $n$-manifold is pointwise conformal to a metric with constant scalar curvature (by the solution of the Yamabe problem). Does this also hold for $n=2$ and/or specifically $S^2$?
No. Conformally equivalent metrics determine the same angles, but not all metrics on $S^2$ (or any given manifold of dimension $> 2$, for that matter) determine the same angles.
You can deduce this in the $S^2$ case from what you already know: For a metric $(S^2, g)$, the Uniformization Theorem guarantees that there is a diffeomorphism $\phi : S^2 \to S^2$ such that $\phi^* g = \lambda^2 g_0$ for some smooth function $\lambda : S^2 \to \Bbb R^+$. Pulling back by $(\phi^{-1})^*$ and rearranging gives that $\lambda^{-2} g = (\phi^{-1})^* g_0,$ but the l.h.s. is conformal to $g$ and the r.h.s. has constant scalar curvature.