Let's say a unit vector $\overrightarrow{A}(t)$ in N dimensions is continuously rotated around the origin. It may for example trace out a circle in 3 dimensions or some sort of spiral in higher dimensions.
Can we derive the (infinitessimal) rotation generator $M$ that is rotating the vector over the path $\overrightarrow{A}(t)$? We would have:
$$A(t) = e^{t M } A(0)$$
where $M=-M^T$ is the rotation generator. I wonder if there is a general formula? Perhaps as an infinite series?
Since $M$ has $N(N-1)/2$ components in $N$ dimensions, I suspect that in 3 dimensions one would need at least first and second derivatives of $A(t)$ at a point to work out $M$ and more in higher dimensions.
In 2 dimensions it would be easy. The solution would be
$M=\begin{bmatrix} 0 & \theta \\ -\theta & 0 \end{bmatrix}$ with $\theta=\frac{\dot{A}_0(0)}{A_1(0)}=-\frac{\dot{A}_1(0)}{A_0(0)}$
I think in N dimensions you would set up a series of $N(N-1)$ simultaneous equations:
$$\partial^{k+1} A(t) = M \partial^k A(t)$$
for $k=1..N-1$ then you could solve for the components of $M$. Although this gives twice the number of equations we need. If we had $M\subset GL(D)$ this would be easier as the solution would be:
$$ M_{nm} = \sum\limits_{k=0}^{D-1}(\partial^{k} A_m(0))^{-1} \partial^{k+1} A_n(0)$$ where the inverse is a matrix inverse over the matrix with indices $k$ and $m$. Not sure how it would change for $M\subset SO(D)$. There would be some identities such as $|A(t)|^2=1$ and all it's derivatives.
Edit (Clarification due to cases pointed out by John Hughes) There will be certain 'degenerate cases' where it is impossible. For example rotating a 3D vector about it's axis traces a point. Let us ignore those cases! (Or on the other hand, find conditions when it is impossible.)
I assume you define rotation in $\mathbb{R}^n$ as simply an element of $SO(n)$.
$M$ is not unique, unless the image of $A$ spans at least a codimension-1 subspace. In 2 and 3D, this holds as long as $A'\neq 0$, but in 4D (and hence higher dimensions) you can have $\begin{pmatrix}R_\theta\\&R_\psi\end{pmatrix}$ for any $\psi$ and for $A(0)=e_1$ you only see a rotation in the $e_1,e_2$-plane and know nothing about that is going on in the $e_3,e_4$-plane.
So, let's assume $A(t)$ spans a codimension $\leq 1$ subspace. We need to solve $$ A(t)=a_0+e^{it\theta_1}(a_1+ia_2)+e^{-it\theta_2}(a_1-ia_2)+\dots+e^{it\theta_m}(a_{2m-1}+ia_{2m})+e^{-it\theta_m}(a_{2m-1}-ia_{2m}) $$ We have $a_0,a_1,a_2,\dots,a_{2m}$ are mutually orthogonal, $a_1,\dots,a_{2m}$ are nonzero, and $\theta_1,\dots,\theta_m$ are distinct nonzero, in order for $A(t)$ to span a dimension $2m$ (if $a_0=0$) or $2m+1$ (if $a_0\neq 0$) subspace of $\mathbb{R}^n$. Thus there is a unique linear dependence between $A'(0),A''(0),\dots,A^{(2m+1)}(0)$ (which therefore also works for time $t$). The coefficients gives you the $\theta_k$ (or rather, $\theta_k^2$), and then looking at kernels give you the $a_k$'s.
If you know the Frenet-Serret formula for a curve in $\mathbb{R}^n$, it is easier to state the result. Since $e^{tM}$ is a 1-parameter subgroup of isometries and $A(t)=e^{tM}A(0)$, all generalized curvatures are constant. So with respect to a Frenet-Serret frame $$ E_1(t)=\frac{A'(t)}{\lvert-\rvert}, E_2(t)=\frac{E_1'(t)-\dots}{\lvert-\rvert},\dots,E_{2m}=\frac{E_{2m-1}'(t)-\dots}{\lvert-\rvert} $$ we have $$ (E_i'(t))=\underbrace{\lvert A'(t)\rvert\cdot\begin{pmatrix}0&-\kappa_1(t)\\ -\kappa_1(t)&0&\ddots\\ &\ddots&\ddots&\kappa_{2m-1}(t)\\ &&-\kappa_{2m-1}(t)&0 \end{pmatrix}}_{\text{constant skew-symmetric matrix}}(E_i(t)) $$ But $E_i(t)$ is just another orthonormal basis if $\{A'(t)\mid t\in\mathbb{R}\}$ spans the whole space. So just conjugate with a change-of-basis to get $M$. The extension to $a_0\neq 0$ and codimension-1 is obvious.