Clearly if three of the points are collinear and the fourth one is not, then we can figure out which center corresponds to which (through the Euler line). Thus all four centers must be collinear for this to happen. Lemma. The four centers are collinear iff the triangle is isosceles. Proof. The "if" way is trivial, since the altitude to the base of the isosceles triangle is also the angle bisector and the median. (Thus all three of those centers must be collinear.) The "only if" statement can be proven with trilinear coordinates: the incenter, orthocenter, and circumcenter are collinear if $$\begin{vmatrix}1&1&1\\ \cos A &\cos B &\cos C\\ \sec A &\sec B &\sec C\end{vmatrix} = 0.$$Then clearly two of $a, b, c$ must be equal. From here, based on intuition, we need the incenter to be the midpoint of the centroid and the orthocenter. I think the answer is a triangle with sides $\boxed{2 - \sqrt{17} - \sqrt{17}}$ by using GeoGebra (it looks about right), but I don't know how to prove it without bashing the coordinates of each center.