When I researched the Fermat-Dao-Nhi equilateral triangle in preamble before points X(33602) of the Kimberling triangle center. I discovered the general result for polygon as follows:

Let $A_1$, $A_2$,...$A_n$; $B_1$, $B_2$,...$B_n$ be $2n$ points and a regular $n$-gons $P_1P_2...P_n$ in the plane; $k, \ell$ are two real numbers such that

1. $\overrightarrow{A_iB_i}=k\overrightarrow{P_iP_{i+1}}$ for $i=\overline{1,n}$ and $P_{n+1} \equiv P_{1}$;
2. Let $3n$ points $C_i$, $D_i$, $E_i$ in the plane such that $\overrightarrow{C_iD_i}=\ell\overrightarrow{C_iA_i}$ and $\overrightarrow{C_iE_i}=\ell\overrightarrow{C_iB_i}$ for $i=\overline{1,n}$

then $D_1D_2...D_n$ be the regular $n$-gons $\Leftrightarrow$ $E_1E_2...E_n$ be the regular $n$-gons

The result above also is a generalization of the Napoleon theorem. Explanation as follows: We apply the theorem with $n=3$; $A_1$, $A_2$, $A_3$ are the vertices of triangle $\triangle ABC$ respectively; $B_1$, $B_2$, $B_3$ are the vertices of the Fermat triangle of triangle $\triangle ABC$ respectively;

1. Because the property of the Fermat line we have $A_1B_1=A_2B_2=A_3B_3$ and angle of this segments form $60^0$ (Fermat point).

2. Let $C_1$, $C_2$, $C_3$ are the vertices of the medial triangle of $\triangle ABC$ respetively. Let points $D_1$, $D_2$, $D_3$ in the plane such that $\overrightarrow{C_iD_i}=\frac{1}{3}\overrightarrow{C_iA_i}$; Let points $E_1$, $E_2$, $E_3$ in the plane such that $\overrightarrow{C_iE_i}=\frac{1}{3}\overrightarrow{C_iB_i}$ for $i=1, 2, 3$. Easily, we can see that $D_1\equiv D_2 \equiv D_3$ is the centroid of $\triangle ABC$ (a point), and $E_1$, $E_2$, $E_3$ are the centroid of the equilateral constructed on the sides of $\triangle ABC$.

By the theorem, because the centroid of $ABC$ is a point so $E_1E_2E_3$ must be an equilateral triangle.

Question: I conjecture that the result continuing hold for higher dimention with similar states. Is the conjecture correct?