Umm not sure if this works so tell me if I'm wrong Let $I_a$ be $A-$excenter in $ABC$. One can easily observe that $(I,I_a ; A1,A2) =-1$. Now consider an inversion with center $I$ and radius $-2\sqrt{pow(I,(ABC))}$ which swaps $A,I_a$ and it sends $A_1$ to the further intersection of $IA_1$ with $(I_aI_bI_c)$ which is clearly a fix circle and finally it sends $I$ to infinity. Since inversion preserves cross-ratio , we'll get that $A_2^{*}$ is a point on $IA$ with some constant ratio made on $IA_1^{*}I_a^{*}$ . So it moves on a cirlce. So the answer is either a circle or a line but clearly the there's a case where $A_1$ lies on $(ABC)$ which maps $A_2$ to infinity. So the final answer should be a line(and to construct it , just consider 2 cases and connect the lines)

Seriously a P4? Recall the well-known property that the isogonal conjugate of a point on the $A-$ angle bisector of $\triangle ABC$ is the inversion of it wrt $\odot (D, DB)$, where $D$ is midpoint of arc $BC$. Then it's easy to calculate that $IA_2*IM$ is fixed value where ${M}$ is midpoint of $AD$. Because ${M}$ is on $\odot OI$, $A_2$ is on a fixed line.

Generalization: $\triangle {ABC}$ has fixed circumcircle $\Omega$ and fixed inscribed conic $\omega$ with center ${I}$. $A_1$ is the reflection of $A$ about ${I}$, and $A_2$ is isogonally conjugated to $A_1$ wrt $\triangle {ABC}$. Prove that as $\triangle {ABC}$ moves, $A_2$ is on a fixed line.

My proof of the generalization: Let $P, Q$ be the foci of $\omega$ and ${O}$ the center of $\Omega$. Line $AI$ meets $\Omega$ again at ${D}$ and ${L}$ is the midpoint of $A_1A_2$. Let the circumscribed circles of $APD, AQD$ be $\odot O_1, \odot O_2$. Line $PQ$ cuts $\odot O_1, \odot O_2$ again at $N, M$ and the pedals of $\odot O_1, \odot O_2$ on it are $S, R$ respectively. Obviously $P, Q$ is an isogonal conjugate wrt $\triangle {ABC}$. Claim: ${D}$ is the spiral similarity center of $P, Q$ and $A_1, A_2$. It is well-known that the spiral similarity center of two pairs of isogonal conjugates of a triangle is on its circumcircle and that it is the isogonal conjugate of the Newton line of the four points. By homothety at $A_1$ , $IL \parallel AA_2$, combining $AD, AA_2$ are isogonal lines of $\angle BAC$, ${D}$ is the desired spiral similarity center. Note that $APA_1Q$ is a parallelogram, simple angle chasing yields $A_2P, A_2Q$ are tangents of $\odot O_1, \odot O_2$ respectively. By $MI \cdot IQ=AI \cdot ID=NI \cdot IP$, $M, N$ are fixed points and $R, S$ are fixed. Since $O_1)_2$ pass through a fixed point ${O}$, there exists some constant $x, y, \overrightarrow{z}$ such that $x\overrightarrow{O_1S}+y\overrightarrow{O_2R}=\overrightarrow{z}$, which indicates there exists some constant $x', y', z'$ such that $x'\cot \angle{A_2PQ}+y'\cot \angle{A_2QP}=z'$, so $A_2$ is on a fixed line.