**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](https://en.wikipedia.org/wiki/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?



Exist $A_1, A_2,\cdots , A_n$ be $n$ points on a sphere, satisfy if $i-j=l-m$ then $A_iA_j=A_lA_m$
The vertices of a [twisted prism](https://en.wikipedia.org/wiki/Prism_(geometry)#Twisted_prism) will achieve this for any even $n$. Choose $0 < r < 1$. Put
$$A\_k = \left( r \cos \frac{2 \pi k}{n}, r \sin \frac{2 \pi k}{n}, (-1)^k \sqrt{1-r^2} \right) .$$




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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.




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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](https://en.wikipedia.org/wiki/Gramian_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.


