nice. Let $A = a, B = b, C = c$ be points on the unit circle in the complex plane and $P = p$ a point not lying on it(to make $A'B'C'$ a non-degenerated triangle). Denote by $G = g$ the centroid of $A'B'C'$. Note that: $A' = \frac{1}{2}(b+c+p-bc \bar p), B' = \frac{1}{2}(a+c+p-ac \bar p), C' = \frac{1}{2}(a+b+p-ab \bar p)$ $\Rightarrow G = \frac{1}{3} \cdot \frac{1}{2} \cdot (3p + 2(a+b+c) - \bar p(ab + ac + bc))$ Lets prove that this is a constant as the triangle moves on the unit circle. Note that $a+b+c$ is the ortocenter of the triangle, which is always on the circumcenter($0$), so its a constant as $p$ and $\bar p$. A known fact is that iff $ABC$ is equilateral, $a^2 + b^2 + c^2 = ab + ac + bc \Rightarrow (a+b+c)^2 = 3(ab+ac+bc)$, so $ab+ac+bc$ is also constant. Hence, we got that the expression for $G$ is a constant, giving us that $G$ is fixed in an unique point as the triangle moves on the circle.

We claim that it is the midpoint of $OP$ where $O$ is the center of $\omega$. For that, let's fix $ABC$, and vary $P$ on a line perpendicular to $BC$. Then the projections of $P$ onto $AB$ and $AC$ move with the same speed, with equal movements in the direction parallel to $BC$. So the centroid of $A'B'C'$ is on a line perpendicular to $BC$. When $P$ is on $BC$, the centroid of $A'B'C' = \frac{A' + C' + B}{3} + \frac{A' + B' + C'}{3} - \frac{B + C - A'}{3}$. The first coordinate corresponds to the centroid of $BPC'$, and so on. This gives that the distance of the centroid from the perpendicular bisector of $BC$ is half the distance from $P$ to the perpendicular bisector of $BC$. Doing this for all sides gives that the required centroid is indeed the midpoint of $OP$.

The complex bash computation is even clearer if one sets $A = 1$, $B = \omega$, $C = \omega^2$ where $\omega = \frac{-1+i\sqrt{3}}{2}$.