Dear Alex,
I generalize the fact as following:

Given triangle ABC with fixed point (not infinite) U with barycentrics (u:v:w) and variable point X.

Take any point A1 on Cevian line XA
Parallel line from A1 with UC cuts XB at B1
Parallel line from B1 with UA cuts XC at C1
Parallel line from C1 with UB cuts XA at A2
Parallel line from A2 with UC cuts XB at B2
Parallel line from B2 with UA cuts XC at C2
Parallel line from C2 with UB cuts XA at A3

The locus of X such that A3=A1 with any A1 on XA is union of:
1. One circumconic:
w*(u + v)*x*y + u*(v + w)*y*z + v*(w + u)*z*x = 0
2. One circumcubic:
CyclicSum=0

Some special triangle centers on this locus:

U = Incenter, X= (Incenter, Orthocenter, Nagel Point, Spieker Center... )
U = Centroid, X= (Centroid, it is special case: there is only Centroid is the triangle center on the locus)
U = Circumcenter, X = (Circumcenter, Orthocenter, Nine Point Center... )
U = Orthocenter, X = (Orthocenter, Incenter, Circumcenter... )
U = Nine Point Center, X = (Nine Point Center, Circumcenter, Midpoint of Nine Point Center and Circumcenter... )
U = Symmedian Point, X = (Symmedian Point... )
U = Gergonne Point, X = (Gergonne Point... )
U = Nagel Point, X = (Nagel Point, Incenter,... )

I propose one problem when U = Incenter and X = Nagel Point: in this case, A1,B1, C1, A2, B2, C2 are not concyclic, but they bound another six concylic points.

We construct six following intersections:

A12 = A1B1, A2C1
B12 = B1C1, B2A2
C12 = C1A2, C2B2

A21 = A2B2, A1C2
B21 = B2C2, B1A1
C21 = C2A1, C1B1
Please prove that six points A12, B12, C12, A21, B21, C21 are concyclic.
This problem seems very hard!

Best regards,
Bui Quang Tuan

>
>I will collect some special cases with simple points so may
>be we can find interesting elementar proof.
>
>Best regards,
>Bui Quang Tuan