Take cendroid G of n poins ,from the leibniz theorem these n points are cyclic with center G And answers are 1,2,3,4(3-4 rectangle),6(3-5-3-5-3-5 equiangular hexagon)

We claim that the only possible values of $n$ are $1, 2, 3, 4, 6.$ For $n = 1, 2$, the constructions are trivial. For $n = 3$ take the vertices of an equilateral triangle of side length $1.$ For $n = 4$ take the vertices of a $3 \times 4$ rectangle. For $n = 6$, first draw an equilateral triangle $V_1V_3V_5$ of side length $7.$ Rotate it about its center to a triangle $v_1'v_3'V_5'$ so that $V_1V_1' = V_3V_3' = V_5V_5' = 3.$ Then, $V_1, V_3, V_5, V_1', V_3', V_5'$ are a suitable choice of six points. Let's now show that no other values of $n$ work. Lemma 1. Any set of points satisfying the property in the question must be concyclic. Proof. This is in the above link. $\blacksquare$ If $n$ is odd, then observe that each distance must "appear" two times per vertex. This is true because it's impossible to pair up the vertices. As a result, if we consider the smallest distance, it becomes apparent that each vertex is equidistant to its two neighbors on the circle. This implies that the polygon formed by the vertices is regular. This easily gives that only $1, 3$ work for $n$ odd. If $n$ is even, then observe we claim that it must be a union of two concyclic $\frac{n}{2}-$gons. If $n = 2,4$, this is obvious, so suppose that $n>4.$ Indeed, consider the smallest distance once again. If it appears twice per vertex, then our polygon is just an $n-$gon, which is clearly a union of two concyclic $\frac{n}{2}-$gons. Now, suppose that it appears only once. It's possible to label the vertices of the $n-$gon with $v_1, v_2, \cdots, v_n$ in clockwise order so that $v_{2i-1}v_{2i}$ is the minimal distance, for each $1 \le i \le \frac{n}{2}$. Now, observe that the distances from $v_1$ to $v_1, v_3, \cdots, v_{n-1}$ are the same as the distances from $v_2$ to $v_2, v_4, \cdots, v_n.$ Consider the minimal distance from $v_1$ to some even vertex (even means even-indexed). By the previous observation, this must equal the minimal distance from $v_2$ to an odd vertex. It's easy to show this implies that $v_1, v_n$ are the same distance apart as $v_2, v_3.$ Using $n-1$ similar arguments, the result is hence proven. Now, it remains to notice that for $n > 3,$ $n-$gons cannot have all sides and diagonals of integral length because $\cos \frac{2 \pi}{n}$ is irrational. This shows that even $n \ge 8$ do not work, and so we've proven that $1, 2, 3, 4, 6$ are the only possibilities. $\square$