INTRODUCTION. I am teaching a course in Harmonic Analysis. In class, very often I find myself stressing out the fundamental property that the functions
$$
e_n(x)=\exp(2\pi i n x), \quad \text{where }\quad x\in \mathbb R / \mathbb Z$$
are simultaneous eigenfunctions of all translations. By this I mean that, defining
$$\tag{1}
\tau_y f(x):=f(x-y), \quad \text{for }x, y\in\mathbb R/ \mathbb Z$$$$\label{1}\tag{1}
\tau_y f(x):=f(x-y), \quad \text{for }x, y\in\mathbb R/ \mathbb Z$$
it holds that
$$\tag{2}
\tau_y e_n=\lambda_{y,n}e_n, \quad \text{for all }y\in\mathbb R/\mathbb Z.$$$$\label{2}\tag{2}
\tau_y e_n=\lambda_{y,n}e_n, \quad \text{for all }y\in\mathbb R/\mathbb Z.$$
QUESTION. Let us now replace $\mathbb R/\mathbb Z$ with the sphere $\mathbb S^2$. By analogy with (1), define $$ \tau_A f(\omega):=f(A^{-1}\omega), \quad \text{for }A\in \mathrm{SO}(3),\ \omega\in\mathbb S^2.$$ Is there a non-constant $f\in L^2(\mathbb S^2)$ such that $$\tag{3} \tau_Af=\lambda_A f, \quad \text{for all }A\in \mathrm{SO}(3)?$$$$\label{3}\tag{3} \tau_Af=\lambda_A f, \quad \text{for all }A\in \mathrm{SO}(3)?$$
REMARKS. From a more general standpoint, the simultaneous diagonalization property (\eqref{2)} of $e_n$ is a common feature of all locally compact abelian groups. Indeed, if $G$ denotes such a group, then we can define $$ \tau_g f(h):=f(g^{-1}h), \quad g,h\in G,\ f\in L^2(G), $$ yielding a unitary representation of $G$. And since $G$ is abelian, all irreducible subrepresentations must be one-dimensional, or in other words, they must be eigenspaces of all $\tau_g$ simultaneously. This is precisely the diagonalization phenomenon observed above in the case $G=\mathbb R / \mathbb Z$.
The present question occured to me while I was trying to find a concrete, pedagogical example of the failure of this simultaneous diagonalization in the non-abelian case.
ANSWER. I can actually provide an answer to the question, which unfortunately I find unsatisfactory for reasons that I will explain below. The answer is the following.
We claim that no non-constant function $f$ satisfying (\eqref{3)} exists. To prove this, introduce the projectors $$ P_{n} f(\omega):=\int_{\mathbb S^2} f(\nu)L_n(\omega\cdot \nu)\ d\sigma(\nu),$$ where $L_n$ is the Legendre polynomial of degree $n$, with an appropriate normalization (the exact value of which is irrelevant here). For each $n\in \mathbb N_{\ge 0}$, $P_n f$ is a spherical harmonic of degree $n$. It is clear that $P_n$ commutes with $\tau_A$ for all $A\in\mathrm{SO}(3)$, so by (\eqref{3)} $$ \tau_A\left( P_n f\right) = P_n \tau_Af= \lambda_AP_n f. $$ So, if a non-constant $f$ satisfying the property (\eqref{3)} existed, we could find a spherical harmonic $Y_n=P_n f$ of degree $n>0$ that satisfies the same property. But this cannot be, for in that case $Y_n$ would span an irreducible subrepresentation of the space $\mathbb{Y}_n$ of spherical harmonics of degree $n$ and we know that $\mathbb{Y}_n$ is irreducible.
The only possibility is that $f$ is a constant, for in that case $P_n f=0$ except for $n=0$. $\Box$
CONCLUDING REMARK. The answer provided above is unsatisfactory for my teaching purposes, since it is too technical and requires the introduction of several concepts, such as spherical harmonics and irreducible subrepresentations. Ideally, I would like an answer that is geometrical in nature and which directly exposes the role of the non-Abelianity of $SO(3)$.
