This is geometry, not combinatorics. Sketch: Only $n=3$ works. Take a segment of maximal length, it can only be a side of an equilateral triangle, say $ABC$. Then any other point must be in the union of $k_1(A, AB)$, $k_2(B, AB)$, $k_3(C, AB)$ and we get a contradiction in both of the two arising cases: $D \in k_1(A, AB)$ but not in the interior of $ABC$; and $D$ in the interior of $ABC$.

Hello, is this your solution, or from the shortlist?

Pretty much both - one of the shortlist official solutions is around this idea and I also solved this problem myself in this way when I tried it.

$|S| = 3$ only. Consider $m$, the longest segment in $S$. It must be a side of some regular polygon in $S$. If that polygon is not a triangle, then one of its diagonals is in $S$ and is longer than $m$, contradicting the fact that $m$ is the longest segment. Therefore the largest shape is $S$ must be an equilateral triangle (which has no interior diagonals). Let the big triangle be $ABC$, and WLOG let $AB = AC = BC = 3$. Then you can show that the second-largest shape must also be a triangle $XYZ$, with a side length $XY = XZ = YZ = r$ where $2 \leq r \leq 3$. If $r \leq 3$, you can get a contradiction by showing one of $\{AX, AY, AZ, BX, BY, BZ, CX, CY, CZ\} \geq r$. It follows that $r = 3$ and some more geo shows that $XYZ = ABC$ so there can only be a single triangle in $S$.

Only $n=3$ if there exist a regualar polygon $\mathcal{P}$ such that $V(\mathcal{P})>3$,We know for all polygons that has a number of vertices over 3 the longest diagonal is greater than it’s side,so it wont be equal to any of it’s side. But,can not we construct a regular polygon such that it’s side equals to the diagonal of the previous polygon? Well when constructing a such polygon we need to construct a new polygon such that it’s diagonal equals to the side of the second polygon we constructed.And we do that for infinity.$\lambda $