Is there always a nontrivial tangent spherically symmetric vector field on an odd sphere? Let's look at an $n$-dimensional sphere embedded in $\mathbb{R}^{n+1}$. Let's consider a vector field $\vec{E}$ which is:
a. Tangent to the sphere.
b. Spherically symmetric (in the ambient space).
c. It is nontrivial: $\vec{E} \neq 0$ (somewhere).
By the Hairy Ball theorem, if $n$ is even, then $\vec{E}$ vanishes somewhere, and so by spherical symmetry it must vanish everywhere.
For $n=1$, there is such a field: $\vec{E}=a\hat{\theta}$.
Is is true that generally in the odd case we may find such a field? Or is there a condition on $n$?

Per the comments, "spherically symmetric" here means invariant under the set of rotations of $\Bbb R^{n + 1}$ (fixing $0$), i.e., the standard action of $SO(n + 1)$.
For $n > 1$ the answer is no: Pick a point $p$ at which the vector field $E$ does not vanish. Then, there is a nontrivial rotation that fixes $p$ but sends $E_p$ to $-E_p$, hence $E$ is not invariant. (For $n = 1$ the only rotation that fixes $p$ is the identity.)
