The group of orthogonal transformations carries an invariant probability measure. This means that we can average a function over the group in a natural way. In particular, if $f$ is a function on the sphere and $θ$ is some point on the sphere, the average over orthogonal $U$ of the value $f(Uθ)$ is just the average of $f$ on the sphere: averaging over $U$ mimics averaging over the sphere: $$\text{avg}_U f(Uθ) = \int_{S^{n-1}}f (\phi) dσ(\phi)$$
See Pg. $22-23$ of these notes for context.
Is there an easy way to understand/prove this? I'm not able to figure out what it means!
My intuition:
Consider $\theta$ on the sphere $S^{n-1}$. Orthogonal transformations rotate this without changing the length. All orthogonal transformations send this point to someplace on the sphere, and for any two points on the sphere, we can always find an orthogonal transformation that relates them. Hence, averaging over all orthogonal transformations is the same as averaging over the sphere.
Is that correct?
Let $\sigma$ be the usual rotation-invariant surface measure on the sphere $S^{n-1}$. We know that in fact $\sigma$ is the only rotation-invariant measure on the sphere. Now, if we let $m$ denote the rotation-invariant (Haar) measure on the group $SO(n)$, let us define a new measure $\nu$ on the sphere, defined by duality via $$ \int_{S^{n-1}} f \,d\nu := \int_{SO(n)} f(U \theta) \,dm(U) $$ for $f: S^{n-1} \to \mathbb{C}$ continuous, where $\theta \in S^{n-1}$ is a fixed arbitrary point. Then the rotation invariance of $m$ on $SO(n)$ implies that $\nu$ is also a rotation-invariant measure on $S^{n-1}$: $$ \int f \circ M \,d\nu = \int_{SO(n)} f(MU \theta) \,dm(U) = \int_{SO(n)} f(U\theta) \,dm(U) = \int_{S^{n-1}} f \,d\nu $$ for any $M \in SO(n)$.
Therefore we conclude that $\nu$ and $\sigma$ are the same measure, i.e. $$ \int_{S^{n-1}} f \,d\nu = \int_{SO(n)} f(U\theta) \,dm(U) = \int_{S^{n-1}} f \,d\sigma $$