We have a triangle.
We then construct three points outside of the triangle by drawing three equilateral triangles on the sides of the original triangle.
Now we want to do the opposite: from the three points we constructed, we want to construct the original triangle. (With just a ruler and a compass)
How does that construction look like?
My idea was to use the Fermat point by drawing the same construction with the three points (the Fermat point is the same).
Denote the points like on this picture:
Consider the following composition of rotations: $I= R_{C',60^\circ}\circ R_{A',60^\circ}\circ R_{B',60^\circ}$. The classification of isometries says that $I$ is a central reflection, but also $I(A)=A$, so $I$ is the central reflection with respect to $A$: $I=S_A$. Thus $A$ can be constructed (e.g. take arbitrary $X$ and construct $X':=I(X)$; $A$ is the midpoint of $XX'$). In a similar way, by considering suitable compositions of rotations, you can construct $B$ and $C$.