How many pairs of points at least on the unit sphere can be used to determine a rotation matrix in $\mathbb{R}^n$ uniquely? A rotation matrix map the unit sphere $\mathbb{S}^n$ onto itself, we want to identify the rotation matrix $R$ by pairs of points $x,x'\in \mathbb{S}^n$, such that
$$
Rx = x'
$$
for $n = 2$, one pair of points is enough, for the general case, is it true that $n-1$ pair of points are enough?
for $n = 3$, using the axis-angle representation of rotation matrix, the axis of rotation passing through the center of the unit sphere and intersecting the unit sphere with $e$, then $d(e,x) = d(e,x')$, that is $e$ must be on the circle that is the intersection of the unit sphere with the perpendicular plane of the segment $xx'$ passing through its middle point. using another pair of points can found another circle, and the point $e$ is the intersection of the two circles. so two pair of points should be enough.
 A: Yes, it is always possible to uniquely specify the action with $n-1$ points,  and fewer than $n-1$ points will never uniquely specify a rotation. We can also construct the unique rotation from any set of points that do admit exactly one solution.
For some positive integers $n$ and $k$, suppose we have sequences $A = \{ x_i \}_{1 \leq i \leq k} \subset S^{n-1}$ and $B = \{ x_i' \}_{1 \leq i \leq k} \subset S^{n-1}$, with each sequence containing no repeated vector values. We are interested in the set of all rotation matrices satisfying $Rx_i = x_i'$ for each $1 \leq i \leq k$.
A rotation matrix must preserve distances, or equivalently inner products, meaning for all index pairs $i,j$,
$$\|x_i-x_j\| = \|x_i'-x_j'\| \iff \langle x_i, x_j \rangle$$
If that's not true, there is no solution. For the rest, we'll assume these conditions are always true.
Let $d_A = \dim \mathrm{span}(A)$. Given $A$ and $B$, let $R_{AB}$ be the linear transformation $R_{AB}: \mathrm{span}(A) \to \mathbb{R}^n$ defined by
$$ R_{AB} \sum_{i=1}^k a_i x_i = \sum_{i=1}^k a_i x_i' $$
Note $R_{AB}\, x_i = x_i'$, and the image of $R_{AB}$ is exactly $\mathrm{span}(B)$.
Let $\{b_i\}_{1 \leq i \leq d_A}$ be any orthonormal basis of $\mathrm{span}(A)$. So for some real numbers $\beta_{ij}$ with $1 \leq i$ and $1 \leq j \leq d_A$,
$$ b_i = \sum_{j=1}^k \beta_{ij} x_j $$
$$ \langle b_i, b_j \rangle = \delta_{ij} $$
where $\delta_{ij}$ is the Kronecker delta.
For any two (possibly equal) indices $1 \leq i,j \leq d_s$,
$$ \begin{align*}
\langle R_{AB} b_i, R_{AB} b_j \rangle &= \left\langle R_{AB} \sum_{u=1}^k \beta_{iu} x_u, R_{AB} \sum_{v=1}^k \beta_{jv} x_v \right\rangle \\
\langle R_{AB} b_i, R_{AB} b_j \rangle &= \left \langle \sum_{u=1}^k \beta_{iu} R_{AB} x_u, \sum_{v=1}^k \beta_{jv} R_{AB} x_v \right\rangle \\
\langle R_{AB} b_i, R_{AB} b_j \rangle &= \sum_{u=1}^k \sum_{v=1}^k \beta_{iu} \beta_{jv} \langle x_u', x_v' \rangle \\
\langle R_{AB} b_i, R_{AB} b_j \rangle &= \sum_{u=1}^k \sum_{v=1}^k \beta_{iu} \beta_{jv} \langle x_u, x_v \rangle \\
\langle R_{AB} b_i, R_{AB} b_j \rangle &= \left \langle \sum_{u=1}^k \beta_{iu} x_u, \sum_{v=1}^k \beta_{jv} x_v \right\rangle \\
\langle R_{AB} b_i, R_{AB} b_j \rangle &= \langle b_i, b_j \rangle = \delta_{ij}
\end{align*} $$
So the sequence $\{R_{AB} b_i\}_{1 \leq i \leq d_A}$ is also orthonormal. This also implies the image of $R_{AB}$ (which again is $\mathrm{span}(B)$) has the same dimension as its domain: $\dim(\mathrm{span}(B)) = \dim(\mathrm{span}(A)) = d_A$.
The direct sum of any vector subspace and its normal subspace is always the entire space:
$$ \mathrm{span}(A) \oplus \big(\mathrm{span}(A)\big)^\bot = \mathbb{R}^n
= \mathrm{span}(B) \oplus \big(\mathrm{span}(B)\big)^\bot $$
$$ \dim \big(\mathrm{span}(A)\big)^\bot = n-d_A = \dim \big(\mathrm{span}(B)\big)^\bot $$
So if $R_\bot : \big(\mathrm{span}(A)\big)^\bot \to \big(\mathrm{span}(B)\big)^\bot$ is any orthonormal linear transformation on those normal subspaces, we can define a corresponding transformation $R : \mathbb{R}^n \to \mathbb{R}^n$ on the entire domain: Any vector $v \in \mathbb{R}^n$ has a unique representation as $v = v_A + v_\bot$ where $v_A \in \mathrm{span}(A)$ and $v_\bot \in \big(\mathrm{span}(A)\big)^\bot$, so define
$$ R(v) = R(v_A + v_\bot) = R_{AB} v_A + R_\bot v_\bot $$
If we combine an orthonormal basis of $\mathrm{span}(A)$ with an orthonormal basis of $\big(\mathrm{span}(A)\big)^\bot$, then $R$ maps it to another orthonormal basis, since basis elements from the same subspace are mapped by the same orthonormal transformation $R_{AB}$ or $R_\bot$; and a pairing of a basis element from $\mathrm{span}(A)$ and a basis of an element from $\big(\mathrm{span}(A)\big)^\bot$, which are perpendicular, map to an element of $\mathrm{span}(B)$ and an element of $\big(\mathrm{span}(B)\big)^\bot$, which are also perpendicular.
The other way around, any orthonormal transformation on the entire space, $R: \mathbb{R}^n \to \mathbb{R}^n$, uniquely determines values $x_i' = R x_i$ and a transformation $R_\bot : \big(\mathrm{span}(A)\big)^\bot \to \big(\mathrm{span}(B)\big)^\bot$ found just by restricting the domain.
If $k = n-1$, the sets of points preserve distances / inner products, and the vectors in $A$ are linearly independent, then $d_A = n-1$ and $d_\bot = 1$. There are exactly two orthonormal linear transformations $R_\bot : \big(\mathrm{span}(A)\big)^\bot \to \big(\mathrm{span}(B)\big)^\bot$, and they are an additive inverse pair. These correspond to exactly two orthonormal linear transformations $\mathbb{R}^n \to \mathbb{R}^n$. One of these is a rotation ($\det R = +1$) and the other is a reflection ($\det R = -1$). So there is exactly one rotation $R$ satisfying $Rx_i = x_i'$ for every $i$.
If $k < n-1$, then $d_A \leq k < n-1$, and we can augment the sequences with any unit vectors $x_{k+1} \in \big(\mathrm{span}(A)\big)^\bot$ and any $x_{k+1}' \in \big(\mathrm{span}(B)\big)^\bot$ to increase the dimension $d_A$. The augmented lists still have $\langle x_i,x_j \rangle = \langle x_i',x_j' \rangle$ since the new pairings have $\langle x_i,x_{k+1}\rangle = 0 = \langle x_i',x_{k+1}' \rangle$ for $1 \leq i \leq k$. We can repeat that until $d_A = n-1$, and by the result above still have a rotation satisfying $R x_i = x_i'$ for the longer sequences. Since we could choose more than one possible values of $x_{k+1}'$ with the same value of $x_{k+1}$, this gives multiple rotations satisfying the set of just $k < n-1$ requirements.
