 Solution

Suppose such an $f$ exists. Orthogonality and unit lengths makea continuous map into the unit sphere, since its squared norm is one. This is a homotopy from the identity to the antipodal map. Their degrees would coincide, but for even $n\ge 2$ they are respectively $1$ and $(-1{)}^{n+1}=-1$, a contradiction. Thus no such continuous map exists for even $n$. At $n=0$ there is not even a unit vector orthogonal to either point of $S^0\subset \mathbb{R}$, so nonexistence holds there as well. This is a sphere form of the Hairy ball theorem.