When is $\left\{\frac{1}{2}x_1+\frac{1}{2}x_2\right\}=\left\{\frac{1}{3}x_3+\frac{1}{3}x_4+\frac{1}{3}x_5\right\}$

Question

Let $X$ be a non convex subset of $\mathbb{R}^n$. Is it possible that the sets $$S_2=\left\{\dfrac{1}{2}x_1+\dfrac{1}{2}x_2,x_i\in X\right\}$$ and $$S_3=\left\{\dfrac{1}{3}x_3+\dfrac{1}{3}x_4+\dfrac{1}{3}x_5,x_i\in X\right\}$$ are equal? If yes is there any characterization of these sets $X$?

Answer 1

If two different points $x,y\in X$, then $$\frac12x+\frac12x=\frac13y+\frac13y+\frac13y$$ that is $$x=y$$ a contradiction.

Then $X$ is empty or has only one point, so it is hardly non convex.