Exist $A_1, A_2,\cdots , A_n$ be $n$ points on a sphere, satisfy if $i-j=l-m$ then $A_iA_j=A_lA_m$ Conjecture: Let $n\geq4$. Is there a set of $n$ non-planar point $A_1, A_2,\cdots , A_n$ be $n$ on a sphere (three-dimensional space) satisfying the conditions: if $i-j=l-m$ then $d(A_i, A_j) = d(A_l, A_m)$ for all $i, j, l, m=1, 2, \cdots, n$? Could you give a proof, or a reference, or a counterexamples?
 A: The vertices of a twisted prism will achieve this for any even $n$. Choose $0 < r < 1$. Put
$$A_k = \left( r \cos \frac{2 \pi k}{n}, r \sin \frac{2 \pi k}{n}, (-1)^k \sqrt{1-r^2} \right) .$$

YCor points out in comments that, if we want $i-j \bmod n$ to determine $d(A_i, A_j)$, then this is the only solution. If we only want $i-j$ to determine $d(A_i, A_j)$, then I don't know if there are more solutions, but I would expect so. (To see the difference: When $n=4$, I know we are requiring $d(A_1, A_2) = d(A_2, A_3) = d(A_3, A_4)$, but I don't know whether the OP intended to impose this distance to be the same as $d(A_4, A_1)$.) The point of this comment is to spell out YCor's observation.
Let the cyclic group $C_n$ of order $n$ act on the $A_i$ by shifting the indices. An isometry of a finite subset of $\mathbb{R}^3$ always extends to an isometry of $\mathbb{R}^3$, and uniquely so if that set is not coplanar. So we have a $C_n$ action on $\mathbb{R}^3$ by isometries. This action must preserve the center of gravity of the $\mathbb{R}^3$, and we may translate that center to the origin, so that $C_n$ acts by orthogonal matrices.
The irreducible representations of $C_n$ over $\mathbb{R}$ are all $1$ or $2$ dimensional, so either $\mathbb{R}^3$ is the sum of three one dimensional irreps, or a two dimensional irrep and a one dimensional irrep, and $\{ A_i \}$ is an orbit in the action. It is easy to check that, if $\mathbb{R}^3$ is a sum of one dimensional reps, then there aren't any $n$-point orbits (for $n>2$). 
If $\mathbb{R}^3 \cong W \oplus 1$ where $W$ is a two dimensional rotation representation and $1$ is the trivial representation, then every orbit lies in a plane and we lose again.
The only possibility is that $n$ is even and $\mathbb{R}^3 \cong W \oplus \epsilon$  where $W$ is a rotation representation and $\epsilon$ is the representation $k \mapsto (-1)^k$. The orbits in this case give the solution above.

If we don't require subtraction to wrap around modulo $n$, then there are more solutions for $n=4$, and I expect for all even $n$. Consider the matrix
$$T = \begin{pmatrix} 1 & x_1 & x_2 & x_3 \\ x_1 & 1 & x_1 & x_2 \\ x_2 & x_1 & 1 & x_1 \\ x_3 & x_2 & x_1 & 1 \end{pmatrix}.$$
If $T$ has signature $+++0$, then we can realize $T$ as $A A^T$ for some $4 \times 3$ matrix. Letting $A_1$, $A_2$, $A_3$, $A_4$ be the rows of $A$, we have $A_i \cdot A_i = 1$ (so the $A_i$ lie on the unit sphere) and $A_i \cdot A_j = x_{|i-j|}$ (so the distance $d(A_i, A_j) = \sqrt{2-2 A_i \cdot A_j}$ depends only on $i-j$.) So it is enough to find matrices $T$ of this form with the required signature. This technique of building a point configuration by finding a positive semidefinite matrix of inner products is called using the Gram matrix.
Now, one such matrix is $x_1=x_2=x_3 = -1/3$ (the regular tetrahedron, also known as an antiprism with $4$ vertices.) But, computing the gradent of $\det T$ at this point, we find $\nabla \det T = (32/9, 64/27, 32/27) \neq 0$. So, by the implicit function theorem, there is a whole surface of deformations $\det T=0$ near this point and, sufficiently near this point, they will all be positive semidefinite. The antiprisms form a one dimensional family (where $x_1=x_3$), so there are many solutions which are not antiprisms. 
I expect you could make such a deformation theoretic argument work for any even $n$, and deleting $A_n$ from any even example gives an odd example.
