I "invented" a determinant $|(1,1,1),(a^2+d^2,b^2+e^2,c^2+f^2),(ad,be,cf)|$ where $a\dots f$ are the squared lengths in a quadrangle ($a,d$ opposite), resulting from a triangle plus some center. Only today I realized 1) the formulation as determinant 2) $a+d$ etc. completely suffices to get what I want, resulting in the Neuberg cubic.
What I wanted, of course, were triangle centers with the property $P$: "If $D$ is the center of $ABC$, $A$ is the same center of $BCD$". Clearly the symmetry of the Neuberg determinant guarantees that being zero is invariant under any permutation of $A,B,C,D$. Likewise, for example, the circumcircle of $ABC$ is invariant and consequently the intersection $X74$ of Neuberg cubic and circumcircle has property $P$. The orthocenter $H=X4$ also pops up repeatedly.
Has property $P$ been studied specifically? Functions in the lengths could be symmetrized with Schur functions, giving other invariant curves. The problem also could be formulated directly as (although horrible) iterated function with trilinears. A link would be welcome. The overlap with this question (or better, the answer with shifted focus) suggests a no.
