Derivative of a unit vector

Here is my interpretation of your question: For $n\ge 3$ you want to have a function $E: S^{n-1}\to {\mathbb R}^n$ (where $S^{n-1}$ is the unit sphere centered at the origin in ${\mathbb R}^n$) such that for every smooth nowhere vanishing function $$ r: {\mathbb R}\to {\mathbb R}^n, $$ for every $t\in {\mathbb R}$, the time derivative $r'(t)$ is a scalar multiple of $E(\hat{r}(t))$.

You also want to have an explicit function $E$, but this does not matter: Such a function simply does not exist (as soon as $n\ge 3$).

The reason is that if $E$ existed then for every unit vector $u\in S^{n-1}$ we had a plane $P(u)\subset {\mathbb R}^n$ (maybe a line in some cases) spanned by the vectors $u$ and $E(u)$. For simplicity, I will work with functions $r: {\mathbb R}\to S^{n-1}$ so there is no need to normalize. For such functions you would require that $r'(t)\in P(r(t))$ for all $t$. But, given any unit vector $u\in S^{n-1}$, it is easy to find a function $r$ as above, such that $r(0)=u$ but $r'(0)\notin P(u)$. For instance, take a function parameterizing a unit circle passing through $u$ and orthogonal to $P(u)$. (This, of course, does not work if $n=2$.) The precise example will depend on what the map $E$ is, but, assuming for concreteness that $u=e_1$ and $E(u)=e_n$ (which one can always achieve by choosing suitable Cartesian coordinates once $E$ is given), then I would take
$$ r(t)= (\cos(t), \sin(t), 0,...,0). $$

Of course, if you want $E$ to depend on more data, namely be a function on the tangent bundle of ${\mathbb R}^n$, then such $E$ does exist, namely, it is the function given by your equation (3).