Conditions for when two metrics are conformally diffeomorphic? Let $M$ be a smooth manifold. For a given pair of Riemannian metrics $g,h$ on $M$, are there any known conditions which imply $g,h$ are  conformally diffeomorphic?
Are there any necessary conditions? What about conformal immersions?
In dimensions $2$, since every metric is conformally flat, locally all metrics are conformally equivalent by diffeomorphisms; the interesting question is the global one.
In higher dimensions, there are obstructions to conformal flatness, so even locally not every two metrics are equivalent.
 A: If dimension is greater than $2$ then the space of conformal classes of Riemannian metrics (modulo the group of diffeomorphisms) is infinite dimensional and there is no reasonable set of invariants which can determine the conformal class. To see that the quotient  is infinite dimensional first note that the space of bounded open contractible domains in $R^n, n>2$, (with smooth boundary) is infinite dimensional; the same if you consider such domains modulo the group of Moebius transformations. Now, given such a domain $D$ and a smooth manifold $M$ of dimension $n$, embed the closure of $D$ smoothly in $M$. Next, extend the flat metric from $D$ to the rest of $M$ to a Riemannian metric $g$ such that $D$ is the maximal open subset in $M$ where $g$ is conformally flat. (For this, a generic extension will do the job.) 
In the case of surfaces, the situation  is of course very different. A conformal structure on a surface is determined by finitely many parameters. For instance, you can use the extremal length of a certain collection of simple loops (if I remember the number correctly, of cardinality $9p-9$ where $p$ is the genus). If the surface is connected, noncompact and has infinite topological type then the dimension of the moduli space is again infinite.
Edit. As for necessary conditions, in dimensions at least 4, conformally diffeomorphic metrics have isomorphic Weyl tensors. More precisely, a conformal diffeomorphism $f:(M,g)\to (N,h)$  sends Weyl tensor of $g$ to the Weyl tensor of $h$. An interesting side question is if the converse holds as well:
Suppose that $(M,g), (N,h)$ are nowhere conformally flat (i.e. contain no nonempty conformally flat open subsets) Riemannian manifolds of dimension at least 4. Suppose that a diffeomorphism $f:(M,g)\to (N,h)$ sends the Weyl tensor of $g$ to the one of $h$. Is $f$ conformal? 
This is formally similar to Kulkarni's theorem about isometries.
A: The obstruction to conformal flatness is measured by the Weyl curvature tensor.
A: According to Wikipedia, a 3-dimensional manifold is conformally flat if and only if its Cotton tensor vanishes:
https://en.wikipedia.org/wiki/Cotton_tensor
