Not only do the centroids of $\triangle G_1 G_2 G_3$ and $\triangle G_4 G_5 G_6$ coincide with each other, they coincide with the centroid of $\triangle ABC$. In fact, the "$G$" triangles are themselves equilateral, and they have a name: (inner and outer) Napoleon Triangles.
Note that we can "construct" $G_1$ by rotating $\overrightarrow{AB}$ by $30^\circ$ about $A$ and then scaling by $\frac{\sqrt{3}}{2}\cdot \frac{2}{3} = \frac{\sqrt{3}}{3}$. Writing $R$ for the rotation operator, and $k$ for the scale factor, we have $G_1 = A + k\;R\;(B-A) \qquad G_2 = B + k\;R\;(C-B) \qquad G_3 = C + k\;R\;(A-C)$ (In $\mathbb{R}^2$, take $R$ to be the appropriate rotation matrix; in the complex plane, take $R = e^{i\pi/6}$ (or $e^{-i\pi/6}$, if we should be rotating clockwise).) Then, answering the question is easy: $\begin{align} \text{centroid of } \triangle G_1 G_2 G_3 &= \frac{1}{3}\left( G_1 + G_2 + G_3 \right) \\[6pt] &=\frac{1}{3}\left( A + B + C + k R\left(\; B - A + C - B + A - C \;\right)\;\right) \\[6pt] &= \frac{1}{3}\left( A + B + C \right) \\[4pt] &= \text{ centroid of } \triangle ABC \end{align}$
The same goes for $\triangle G_4 G_5 G_6$, except that we rotate the vectors in the opposite direction.
Proving that the "$G$" triangles are equilateral is left as an exercise for the reader. (One might notice that the specific angle for $R$, and the specific value of $k$, are actually irrelevant to the proof of the centroid relation, since the rotated terms cancel-out. They are, however, key to guaranteeing the equilateral property.)