Label the unknown triangle $\triangle ABC$ with centroid $G$ and symmedian point $K.$ $AG,BG,CG$ cut the circumcircle of $\triangle ABC$ again at the known points $D,E,F.$ $U$ is the centroid of $\triangle DEF$ and $G_1,G_2,G_3$ are the orthogonal projections of $G$ on $EF,FD,DE.$ From the cyclic quadrilaterals $GG_1EG_3$ and $GG_1FG_2,$ we obtain $\angle GG_1G_2=\angle CFD=\angle CAG=\angle KAB.$ Analogously, we have $\angle GG_2G_1=\angle KBA$ $\Longrightarrow$ $\triangle GG_1G_2 \sim \triangle KAB.$ Likewise, $\triangle GG_1G_3 \sim \triangle KAC$ $\Longrightarrow$ $\triangle ABC \cup K \sim \triangle G_1G_2G_3 \cup G,$ i.e. $G$ is the symmedian point of $\triangle G_1G_2G_3.$ According to the topic symmedian point, the circumcenter of any triangle is the centroid of the antipedal triangle of its symmedian point. So we deduce that $U$ is the center of the pedal circle of $G$ WRT $\triangle DEF$ $\Longrightarrow$ $\odot(G_1G_2G_3)$ is the pedal circle of the conic $\mathcal{U}$ with center $U$ tangent to $EF,$ $FD,$ $DE,$ i.e. the Steiner inellipse $\mathcal{U}$ of $\triangle DEF$ tangent to its sides through their midpoints. Thus, the possible points $G$ are the foci of $\mathcal{U}.$ The foci of $\mathcal{U}$ are indeed constructible using ruler and compass by defining an appropriate homology that takes $\mathcal{U}$ into a circle, e.g. see the exercise Construction of conic. Once the foci of $\mathcal{U}$ are constructed, then their circumcevian triangles WRT $\triangle DEF$ determine the two possible $\triangle ABC.$