cv.complex-variables
In dimension 2 we know by the Riemann mapping theorem that any simply connected domain ( $\neq \mathbb{R}^{2}$) can be mapped bijectively to the unit disk with a function that preserves angles between curves, ie is conformal.
I have read the claim that conformal maps in higher dimensions are pretty boring but does anyone know a proof or even a intuitive argument that conformal maps in higher dimensions are trivial?
A quick comment: I assume you want $U$ to be "non-trivial" i.e. not equal to $\mathbb{C}$ itself; if it were, then the collection of such maps should be infinite dimensional (in particular, it would contain every polynomial).
So assume that $U$ is non-trivial. I'll also assume that $U$ is simply connected, though I'm pretty sure that you can do away with this assumption. Thus $U$ is biholomorphic to the unit disc in $\mathbb{C}$, so we will assume it is the unit disc.
The holomorphic self-maps of the unit disc contain the group $G = PSL_2(\mathbb{R})$ (this is its group of automorphisms, actually). This is a real 3-manifold, so if you restrict yourself to biholomorphisms, you're good.
However, it also contains the maps $z \mapsto z^k$, and so all conjugates of these maps by $G$. There might be something more you can say about this, but I'm not at the moment sure what.
You may want to look at Don Marshall's Zipper algorithm: https://sites.math.washington.edu/~marshall/zipper.html
Added in Edit by T. Banakh. This Zipper algorithm yields the following image of the conformal map of the unit disk to an oak leaf.
Many thanks to Prof. Donald E. Marshall for producing this image (which I post here with his permission).
Best Answer
I think you're looking for Liouville's theorem. This theorem states that for $n >2$, if $V_1,V_2 \subset \mathbb{R}^n$ are open subsets and $f : V_1 \rightarrow V_2$ is a smooth conformal map, then $f$ is the restriction of a higher-dimensional analogue of a Mobius transformation.
By the way, observe that there are no assumptions on the topology of the $V_i$ -- they don't have to be simply-connected, etc.
EDIT : I'm updating this ancient answer to link to a blog post by Danny Calegari which contains a sketch of a beautifully geometric argument for Liouville's theorem.