Let $\Omega$ be a smooth bounded domain of $\mathbb{S}^n$, the unit $n$ sphere centered at the origin of $\mathbb{R}^{n+1}$, and consider the functions $$L_{ij}u=x_{i}\frac{\partial u}{\partial x_j}-x_{j}\frac{\partial u}{\partial x_i}\hspace{0.5cm} i,j=1\ldots n+1$$ where $x_i$'s are the Cartesian coordinates for $\mathbb{R}^{n+1}$.
(P) If all these functions $L_{ij}u$ are identically $0$ in $\Omega$, this would imply that $u$ is radially symmetric in $\Omega$ and therefore that $\Omega$ is a ball.
Is (P) is true?
Sorry for the late answer ... First let us discuss the symmetry point. Consider a function $f$ on $\mathbb{R}^{n+1}$ and let $R_{ij}(\varphi)$ be operators, realising rotations in the $x_i-x_j$-plane, i.e. $$ (R_{ij}(\varphi)f)(x_1,\dots,x_{n+1})=f(x_1,\dots,x_i \cos\varphi-x_j\sin\varphi,\dots,x_i\sin\varphi+x_j\cos\varphi,\dots,x_{n+1}) $$ Then we have $$ \frac{d}{d\varphi}\bigg|_{\varphi=0}R_{ij}(\varphi)f=L_{ij}f. $$ Each $R_{ij}$ is a representation of $SO(2)=\mathbb{S}^1$ and by the above relation, $L_{ij}$ is a "infinitesimal rotation", i.e. a representation of the Lie algebra $\mathfrak{so}(2)$. Since $SO(2)$ is simply connected, each finite rotation can be obtained by "integrating the infinitesimal one", so $L_{ij}f=0$ is indeed equivalent to $f$ being invariant under $R_{ij}(\varphi)$ for each $\varphi$.
Next, note that each rotation of the whole space, i.e. each element of $SO(n+1)$, can be composed by these $R_{ij}$. This shows that if we have $L_{ij}f=0$ for all $i,j,=1,\dots,N+1$, then $f$ is invariant under the action of $SO(n+1)$.
Back to your question: If we have such a function $f$, then it must be constant on any $n$-sphere, since the sphere itself is invariant under the action of $SO(n+1)$. In other words: $f(x)$ only depends on the norm $|x|$.
But you are interested in functions on the sphere. The $L_{ij}$ can be properly pulled back to differential operators on the sphere. This can be explicitely seen by writing them in spherical coordinates, where we see that they don't depend on derivatives in the radial coordinate $r$. However, the statement is still valid, $SO(2)=\mathbb S^1$ now acts in the obvious way on the sphere. Hence, every function $u$ on the sphere which fulfills $L_{ij}u=0$ for all $i,j$ is constant.
What if we only consider functions defined on domains $\Omega\subset \mathbb S^{n+1}$? Then, the local statement $L_{ij}u=0$ can still be imposed, no matter what the properties of $\Omega$ are. The integration of the this equation, i.e. the behaviour of $u$ under rotations is more subtle since $R_{ij}(\varphi)u$ is only defined if the rotation leaves $\Omega$ invariant.
However, we can also argue point-wise. Assume that $\Omega$ is path-connected. Let $x,y\in\Omega$ then the connecting path can chosen to be a result of a sequence of rotations (i.e. composed of geodesics). If $L_{ij}u=0$ on $\Omega$, then each of this rotations does not change the value $u(x)$ and hence $u(y)=u(x)$. This shows that $u$ is constant on $\Omega$. If $\Omega$ is not path-connected, then $u$ is constant on each path-component.
Note that the properties of $\Omega$ does not matter!