Not sure what this type of graph is called but it is a type of static balance graph, where if for $n$ dimensions, the vertices satisfy the relation condition 1 (which can be seen as a balance from the origin)
$$\sum_{i=0}^{n} r_i=0$$
as well as an additional relation representing the $n$-simplex polytope geometry with the condition of equal side lengths, condition 2
$$|r_k-r_{k'}|=a \mbox{ for } k \neq k'$$ for some fixed constant $a$, though by some analysis it is possible for it be $a=2sin(\pi/(n+1))$ for unit length vertices.
Defining unit vector $u_l=x_l/|x_l|$, for $n=1$ and $j=0,1$
$$ r_j=\exp(\frac{2 \pi i j}{2})u_0 $$ which satisfies $|r_k-r_{k'}|=2$
For $n=2$ and $j=0,1,2$ $$r_j=\cos(\frac{2\pi j}{3})u_0+\sin(\frac{2\pi j}{3})u_1$$ which satisfies $|r_k-r_{k'}|=\sqrt{3}$
Is there such an expression known for $n>2$? For example, there seems to be something similar, for $n=3$
$$r_j=\cos(\frac{2\pi j}{4})u_0+\sin(\frac{2\pi j}{4})u_1+\cos(\frac{2\pi j}{4})u_2$$
which satisfies condition 1 but not 2. Note also, for $n=1,2$ the formulas can be extended to many $j$, but not to many $j$ in the case of $n=3$
The octahedron does generalize to any dimension, then variously being called hyperoctahedron or crosspolytope. Just as the octahedron has triangular faces, the crosspolytope always will have (regular) simplexial facets.
This observation allows for the most pleasing simplex coordinates possible: simply use (1, 0, 0, …, 0), (0, 1, 0, …, 0), …, (0, 0, 0, …, 1). Obviously those all live within a single affine hyperplane (the span of the chosen facet). In fact, all these points bow to coordinate equation
$$\sum_i x_i = 1$$
You even can calculate the offness of this hyperplane. Simply consider the vector sum, resulting in the hypercube diagonal of length $\sqrt n$. As the vertices of the hypercube come in $n+1$ layers wrt. this hyperbody diagonal, they divide this vector into $n$ equal sized segments.
This now allows for a backshift of this very affine hyperplane, so that its image would become origin-centered again: just subtract this offness from each coordinate. - Or, rescaling everything by the common denominator, you'd get: $(n-1, -1, -1, …, -1)$, $(-1, n-1, -1, …, -1)$, …, $(-1, -1, -1, …, n-1)$, which then clearly bow to the new coordinate equation
$$\sum_i x_i'=0$$
--- rk