Solution:

We observe that, considering the median $AM$, the three segments $AM'$, $M'G$ and $GM$ are equal to one another, since the centroid divides the median into two segments one twice the other and $M'$ is by construction the midpoint of the longer of the two, $AG$. An analogous result holds for the other two medians.

Suppose we have an equilateral triangle, then the three medians are equal to one another, so the points $M$, $M'$, $N$, $N'$, $L$, $L'$ are equidistant from $G$, that is, they lie on a circle centered at $G$.

Suppose now that the six points $M$, $M'$, $N$, $N'$, $L$, $L'$ lie on a circle with center $O$. This point $O$ lies on the perpendicular bisectors of the three segments $MM'$, $NN'$ and $LL'$ and, since $G$ is the midpoint of each of them, $G$ and $O$ lie simultaneously on the perpendicular bisectors of these segments. Since the three perpendicular bisectors are distinct, it follows that $G$ and $O$ must coincide. But then $GM = GN = GL$, being radii of the same circle, and therefore the three medians of the triangle $ABC$, which as stated earlier have length $3GM$, $3GN$, $3GL$, are equal to one another. A triangle in which all the medians are equal is equilateral, since for every pair of equal medians it is isosceles on the two sides to which the medians are relative.