Denote by $A_O$ antipode of $A$ WRT $(ABC)$(with center $O$) and by $A_H$ and $H_A$ intersection point of $A-$altitude in $\triangle ABC$ with $(ABC)$ and $BC$, respectively. We'll consider $4$ cases: If $X\equiv B$, then $Y\equiv B$. If $X\equiv C$, then $Y\equiv C$. If $X\equiv A_O$, then $Y\equiv O$. If $X\equiv A_H$, then note that $\angle H_AA_HQ=\angle ACH_A=\angle H_AA_HB$, thus $Q$ is reflection of $B$ WRT $H_A$, similarly for $C$ and $P$. Thus reflecting $(A_HPQ)$ over $AA_H$ we get $(ABC)$, thus $Y$ is reflection of $O$ WRT $AA_H$. Thus if $O'$ is reflection of $O$ WRT $AA_H$, then $B$, $C$, $O$ and $O'$ lie on circle $\Longleftrightarrow \triangle ABC$ is isosceles with $AB=AC$. Now in that isosceles triangle consider case when $X\equiv A$. Note that $Y\in (BOC)\Longleftrightarrow YB=AB=AC=YC$(because of symmetry) i.e. $Y$ is reflection of $A$ in $BC$, but as reflection of $A$ in $BC$ lies on $(BHC)$ as well it follows that $(BOC)\equiv (BHC)$ i.e. since $AB=AC$ we have $H\equiv O\Longleftrightarrow \triangle ABC$ is equilateral.

I found 3 generalizations: 1) if $ABC$ is arbitrary then the locus of $Y$ is a conic with focus $F_1,F_2$ with center $S$ which lies on $AO$. [asy][asy] import graph; size(13.3724559250421cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -40.28737391388937, xmax = 97.08508201115272, ymin = -39.023809374011826, ymax = 25.452733585043447; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); filldraw((16.198633778972557,5.5851831491241155)--(15.348467135642982,6.853036878625743)--(14.080613406141355,6.002870235296166)--(14.93078004947093,4.7350165057945395)--cycle, evefev, qqwuqq); /* draw figures */ draw(circle((8.034726851046631,8.006078882190646), 12.235783169215052)); draw(circle((26.035883029975647,-12.879065120450552), 21.96334370761289)); draw(shift((13.62214242448413,-12.518172409781231))*rotate(120.1051946201011)*xscale(23.693702738075576)*yscale(10.984768326296559)*unitcircle); draw((2.9154765272611467,14.058561048114557)--(20.266659491480425,8.313021293624193)); draw((20.266659491480425,8.313021293624193)--(43.31289216421736,0.6816657667128443)); draw((20.266659491480425,8.313021293624193)--(14.93078004947093,4.7350165057945395)); draw((14.93078004947093,4.7350165057945395)--(8.620701828524002,0.5037570983246745)); draw((2.9154765272611467,14.058561048114557)--(4.820691264685021,19.81219445902825)); draw((4.820691264685021,19.81219445902825)--(-1.5902983907090278,0.45139299463630034)); draw((4.820691264685021,19.81219445902825)--(14.93078004947093,4.7350165057945395)); draw((14.93078004947093,4.7350165057945395)--(17.73672777906071,0.5505059493530675)); draw((8.620701828524002,0.5037570983246745)--(-1.5902983907090278,0.45139299463630034)); draw((-1.5902983907090278,0.45139299463630034)--(17.73672777906071,0.5505059493530675)); draw((17.73672777906071,0.5505059493530675)--(43.31289216421736,0.6816657667128443)); label("$A$",(3.536526378593938,23.14999990793433),SE*labelscalefactor); label("$B$",(-3.947358072010765,1.561871685036358),SE*labelscalefactor); label("$C$",(17.209007586429454,0.050702709433500005),SE*labelscalefactor); label("$O$",(7.854152023173574,10.916727248292146),SE*labelscalefactor); label("$F_1$",(3.0328033867263136,6.167339039254593),SE*labelscalefactor); label("$F_2$",(23.757406480708568,-26.934457569188968),SE*labelscalefactor); label("$Q$",(8.50179586986052,0.9142278383494189),SE*labelscalefactor); label("$P$",(43.90632615541354,3.0010802332295565),SE*labelscalefactor); label("$X$",(20.66310810209316,11.996133659437046),SE*labelscalefactor); label("$Y$",(26.923665286733634,-10.599440547196167),SE*labelscalefactor); label("$S$",(11.883935958114568,-12.470411659847324),SE*labelscalefactor); draw((4.820691264685021,19.81219445902825)--(13.622142424484128,-12.51817240978123), red); /* dots and labels */ dot((4.820691264685021,19.81219445902825),linewidth(3.pt)); dot((-1.5902983907090278,0.45139299463630034),linewidth(3.pt)); dot((17.73672777906071,0.5505059493530675),linewidth(3.pt)); dot((20.266659491480425,8.313021293624193),linewidth(3.pt)); dot((8.620701828524002,0.5037570983246745),linewidth(3.pt) + uuuuuu); dot((43.31289216421736,0.6816657667128443),linewidth(3.pt) + uuuuuu); dot((26.035883029975647,-12.879065120450552),linewidth(3.pt) + uuuuuu); dot((8.034726851046631,8.006078882190646),linewidth(3.pt) + uuuuuu); dot((2.9154765272611467,14.058561048114557),linewidth(3.pt) + uuuuuu); dot((14.93078004947093,4.7350165057945395),linewidth(3.pt) + uuuuuu); dot((24.15224721550223,-30.679761198983),linewidth(3.pt) + uuuuuu); dot((3.0920376334660293,5.643416379420541),linewidth(3.pt) + uuuuuu); dot((13.622142424484128,-12.51817240978123),linewidth(3.pt) + uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] 2) If there exist 5 distinct points $X_1,X_2,X_3,X_4,X_5$ such that $Y_1,Y_2,Y_3,Y_4,Y_5$ lie on a circle then $ABC$ must be equaliteral. 3) The locus of the second intersection of circumcircles $YPC,YQB$ is a circle.