Let's say we have a set of orthogonal functions (such as spherical harmonics) over some manifold such that (for a simple one dimensional example):
$$\delta_{nm}=\int_{M}Y_{n}(x)Y_{m}(x)dx$$
Now lets consider transforming the setup to a conformally related space. Do we still have that:
$$\delta_{nm}=\int_{M'}Y'_{n}(x')Y'_{m}(x')dx'$$
Given that Conformal transformations preserve angles I expect the answer is a definitive YES. Does anyone know for sure? I'm interested in spherical harmonics on the $S^{3}$ mapped to the Bohr one point compactification of $R^{3}$.
There is some confusion here. The angles that they preserve are angles between vectors drawn on the space $M$ (technically, tangent vectors to $M$). This is totally different from angles between functions defined on $M$, where we use inner product $\langle f, g\rangle =\int_M fg$. It may help to realize that the angle between functions depends only on the measure with which we equip the space $M$, not on its metric.
In any case, the following is true for any diffeomorphism $\varphi:M'\to M$ between Riemannian manifolds (equipped with the Riemannian volume).
If the functions $Y_n$ are orthogonal on $M$, then the functions $Y'_n(y) = Y_n(\varphi(y)) \sqrt{|J_\varphi(y)|}$ are orthogonal on $M'$, where $J_\varphi$ is the Jacobian determinant of $\varphi$. This follows from the change of variable formula, $$ \int_{M'}Y_n'(y)Y_m'(y)\,dy = \int_{M'}Y_n(\varphi(y))Y_m(\varphi(y)) |J_\varphi(y)|\,dy = \int_M Y_n(x)Y_m(x)\,dx = \delta_{mn} $$
If you simply define $Y_n'(y) = Y_n(\varphi(y))$, then the above does not work and $Y_n'$ are not orthogonal in general.
Conformality plays no role here.