Is group homomorphism $F: SO(3) \to \mathbb R-\{0\}$ unique? If we assume that $F$ is smooth, then such $F$ is unique. Under this assumption, the question is equivalent to find all 1-dimensional representation of $SO(3)$, and by considering Lie algebra there is only the trivial one.
But what if we remove the smoothness assumption? I think this question may be related to the fact that there are "nontrivial" $\mathbb Q$-linear map $\mathbb R \to \mathbb R$, which can be constructed by using the axiom of choice.

Yes, it's unique: indeed, the group $\mathrm{SO}(3)$ is perfect: actually every element is a commutator.
Indeed, consider any element $q$: this is a rotation; hence square of another rotation (with same axis) $r$, namely $q=r^2$. Then $r$ and $r^{-1}$ being rotations of the same angle, are conjugate (by any element reversing the axis of $r$): $r^{-1}=srs^{-1}$. So $q=r^2=r(r^{-1})^{-1}=rsr^{-1}s^{-1}$.
