Let $O$ denote the circumcenter of triangle $ASR$. Let $O_1$ denote the image of $O$ under $\chi$. Since $\chi$ takes triangle $OAS$ to triangle $O_1QP$, we have $\triangle O_1QP \sim \triangle OAS \sim \triangle ORA$. Hence, $\phi$ takes triangle $ORA$ to triangle $O_1QP$. This means triangle $XOO_1$ is similar to triangle $XAP$.
Thus, $OX/O_1X = AX/PX = AY/QY = O_1Y/OY$. By the Angle Bisector Theorem, $O_1O$ bisects angle $XO_1Y$. However, $\angle XO_1O = \angle XPA$ and $\angle OO_1Y = \angle AQY$. Therefore, $\angle XPA = \angle AQY$.

We present an alternate way to finish the proof after making the observations in the first paragraph of Solution 1. Let $\theta$ denote the spiral similarity centered at $O$ that takes $R$ to $A$. Consider the images of $P$ and $Q$ under the composition $\chi^{-1} \circ \theta \circ \phi$, which is itself a spiral similarity. We have $P \mapsto A \mapsto S \mapsto P$ and $Q \mapsto A \mapsto R \mapsto Q$, so the composition is a spiral similarity fixing two points and hence the identity.

Let $X_1 = \theta(X)$. The successive images of $X$ under this composition are $X \mapsto X \mapsto X_1 \mapsto X$, which shows that $X_1 = \chi(X)$. We have $\angle XOX_1 = \angle AOS = \angle AXP + \angle AYQ$, so $\angle OXX_1 = 90^\circ - \frac{1}{2}(\angle AXP + \angle AYQ)$. But $X_1 = \chi(X)$, so $\angle OXX_1 = \angle YQA$. Equating the two expressions, we obtain $\angle YQA = 90^\circ - \frac{1}{2}(\angle AXP + \angle AYQ)$. Similarly, we have $\angle APX = 90^\circ - \frac{1}{2}(\angle AXP + \angle AYQ)$. Together, these imply that $\angle XPA = \angle AQY$.