I have been trying to find a way to do a conformal mapping from tetrahedron, specifically a trirectangular tetrahedron, to a spherical sector, but being an engineer with no training in this subject it seems to be a few notches too advanced for me. This 
is an attempt to illustrate the two geometries and how they are placed wrt. eachother.
Most of what I find on conformal mappings is in 2D and formulated in the complex plane, which naturally does not cover my problem. I have found that the mapping must consist of only similarity transformations and inversions, but inversions seem to be mostly applied to spherical shells (e.g stereographic projections) and not to "solid spheres."
I also need the analogous mapping in 2d, i.e. a right isosceles triangle to a disk sector where the base and one leg coincides with the sector rays, but here I assume I can use the mapping from a disc to a square using only a quarter of the disc/square. Is this correct, and when I only need this part is there a formula not involving complex numbers?
Edit: Also, the mapping must be the identity mapping at the centre of the sphere (disk in 2d), so that the (iso-) surfaces of constant radii in the sphere are gradually changing from spherical to planar in the tetrahedron as the radial coordinate increases.
Any map that sends the top "corner" of the "ball sector" in your picture to the top vertex of the tetrahedron in particular sends a right angle to an acute angle, and so cannot be conformal, so there is no map meeting your criteria.
Like Will Jagy says, in 2 dimensions any simply connected domain (other than $\mathbb{C}$) is analytically and hence conformally equivalent to a disc. In short there are lots of conformal maps, and in fact we can make more precise the statement that there is an infinite-dimensional space of conformal maps.
But, the 2-dimensional setting is wholly misleading about conformal geometry in higher dimensions, which is much more rigid, and where there is only a finite-dimensional space of conformal maps.