Suppose we have $n$ points on the 3D unit sphere,
$X = (\pmb{r}_{1}, \pmb{r}_{2}, ..., \pmb{r}_{n})$.
I am interested in knowing to what extent the rotational symmetry of $X$ is determined by the following constraint. Every ordered triple of points ($n^{3}$ of them in total) is related to at least one other triple of points by some rotation about the origin,
$(\pmb{r}_{j_{1}}, \pmb{r}_{j_{2}}, \pmb{r}_{j_{3}}) = (R\pmb{r}_{l_{1}}, R\pmb{r}_{l_{2}}, R\pmb{r}_{l_{3}})$,
where $R$ is a rotation (not the identity) depending on $j_{1}, j_{2}, j_{3}$. The triple $l_{1}, l_{2}, l_{3}$ must be distinct from $j_{1}, j_{2}, j_{3}$, but perhaps only in order.
Clarification:
If the collection of points has rotational symmetry,
$\pmb{r}_{\sigma(j)} = R \pmb{r}_{j}$,
where $R$ is the same for all $j$ and $\sigma$ is a permutation, then the constraint is obviously satisfied. It is thus not hard to find a collection of points satisfying the constraint. What I'm interested in is what one can conclude in the opposite direction: what can one say about the symmetry of the collection of points if the constraint is satisfied (perhaps nothing)?
At one extreme, if the triple is replaced by an $n$-tuple, the collection of points must have some rotational symmetry. One of the constraints is
$(\pmb{r}_{1}, \pmb{r}_{2}, \dots, \pmb{r}_{n}) = (R \pmb{r}_{\sigma(1)}, R \pmb{r}_{\sigma(2)}, \dots R \pmb{r}_{\sigma(n)})$,
which implies
$\pmb{r}_{j} = R \pmb{r}_{\sigma(j)}$.
At the other extreme, where the triple is replaced by a single point, the constraint is obviously satisfied by any collection of points on the sphere. (Every point on the sphere is related to every other point by some rotation about the origin.)
What can one say between these two extremes?
Any advice would be much appreciated.