I want to prove that this action is ergodic. That is, if $\lambda$ is the normalized Lebesgue measure on $S^n$ and $A$ is a measurable set such that for all $g \in SO(n + 1)$ we have $\lambda(gA \Delta A) = 0$, then $\lambda(A) = 0$ or $1$.
I initially thought that this followed from the transitivity of the action (see this question I asked some time ago: Equivalence of definitions of ergodic action). However, it still doesn't seem to bring me anywhere.
My first try was to approximate $A$ by an open set and then use the separability and the compactness of $S^n$, but I run into some problems pretty much straight away.
First, some small calculations show that it is enough to prove that the action is ergodic with respect to the Lebesgue measure restricted to the Borel sets.
Now note that since $G = SO(n + 1)$ and $K = SO(n)$ are compact, the normalized left Haar measure $\mu$ on $G$ induces the pushforward measure $\mu_*$ on the homogenous space $G/K$. This is the measure defined by $\mu_*(A) = \mu(\pi^{-1}(A))$, where $\pi : G \to G/K$ is the canonical projection. But $G/K \cong S^n$, and the Lebesgue measure is the unique rotation-invariant countably additive measure of total measure $1$ on the Borel sets of $S^n$. It follows that $\lambda = \mu_*$.
Now suppose that $\mu_*(gA \Delta A) = 0$. Then a small calculation shows that $\mu(g\pi^{-1}(A) \Delta \pi^{-1}(A)) = 0$. So to show that the action of $G$ on $S^n$ is ergodic, it is enough to show that the action of $G$ on itself is ergodic, with respect to the normalized left Haar measure. This can be proven using, for instance, proposition 8.6 in "Ergodic Theory with a view towards number theory" by Einsiedler and Ward.
Another way to prove this, is using directly proposition 8.3 in the same book. However, the proof of the former is straightforward and even simpler in our case (we only need half of it), while the proof of the latter is more involved and uses previously proven results.