This problem looks interesting, but I don't quite understand it. Can anyone clarify what it means?

Assume that there exists two points $P,Q,$ such that $PA_1,QA_1$ are isogonals WRT $\angle A_5A_1A_2,$ $PA_2,QA_2$ are isogonals WRT $\angle A_1A_2A_3,$ etc. Let $P_1,$ $P_2,$ $P_3,$ $P_4,$ $P_5$ be the projections of $P$ on $A_1A_2,$ $A_2A_3,$ $A_3A_4,$ $A_4A_5,$ $A_5A_1$ and $Q_1,$ $Q_2,$ $Q_3,$ $Q_4,$ $Q_5$ the projections of $Q$ on $A_1A_2,$ $A_2A_3,$ $A_3A_4,$ $A_4A_5,$ $A_5A_1.$ By well known property of isogonals, $P_1,Q_1,P_2,Q_2$ lie on a circle centered at the midpoint $M$ of $\overline {PQ}.$ Similarly, $P_2,Q_2,Q_3,P_3$ lie on a circle centered at $M,$ etc $\Longrightarrow$ $P_1,P_2,P_3,P_4,P_5$ lie on a same circle $(M).$ Hence, $A_1A_2 \perp PP_1,$ $A_2A_3 \perp PP_2,$ $A_3A_4 \perp PP_3,$ $A_4A_5 \perp PP_4$ and $A_5A_1 \perp PP_5$ are tangents of the conic $\mathcal{K}$ with focus $P,Q$ and pedal circle $(M).$ Since $\mathcal{K}$ is unambiguosly defined by the five tangents $A_1A_2,$ $A_2A_3,$ $A_3A_4,$ $A_4A_5,$ $A_5A_1,$ then the pair $P,Q$ is unique.

Call a pair of points $P$, $Q$ that satisfy the conditions of the problem pentagonal isogonal conjugates. Firstly we will show if pentagonal isogonal conjugates $P$, $Q$ exist, then there exists an ellipse with foci $P$, $Q$ that is tangent to all five sides of the pentagon. Let the reflection of $P$ in side $A_iA_{i+1}$, for all $1 \le i \le 5$, be $P_i$. By a well known result of isogonals, $QP_i=QP_{i+1}$ for all $1 \le i \le 5$, so $QP_1=QP_2=QP_3=QP_4=QP_5$. Let $QP_i$ intersect $A_iA_{i+1}$ at $R_i$ for all $1 \le i \le 5$. Then $PR_i+QR_i=P_iR_i+QR_i=QP_i$, and $PR_i$ and $QR_i$ are symmetric with respect to $A_iA_{i+1}$, so there exists an ellipse with foci $P$, $Q$ that is tangent to all five sides of the pentagon. Now for any ellipse tangent to the sides of the pentagon, its foci are pentagonal isongonal conjugates. Since there exists exactly one ellipse tangent to all five sides of the pentagon, we are done.

Quote: Since there exists exactly one ellipse tangent to all five sides of the pentagon, we are done. I would say that you failed to simplify the problem, unless you take an affinement and make the ellipse a circle.

[asy][asy] unitsize(2inches); pair A=dir(115); pair B=dir(215); pair C=dir(-35); real x=0.2; real y=0.65; real z=1-x-y; real a=abs(B-C); real b=abs(A-C); real c=abs(A-B); real t=a*a/x+b*b/y+c*c/z; real xx=(a*a/x)/t; real yy=(b*b/y)/t; real zz=(c*c/z)/t; pair P=x*A+y*B+z*C; pair Q=xx*A+yy*B+zz*C; pair PA=2*foot(P,B,C)-P; pair QA=2*foot(Q,B,C)-Q; pair PB=2*foot(P,A,C)-P; pair QB=2*foot(Q,A,C)-Q; pair PC=2*foot(P,A,B)-P; pair QC=2*foot(Q,A,B)-Q; pair D=extension(P,QA,B,C); pair E=extension(P,QB,A,C); pair F=extension(P,QC,A,B); real angle = degrees(atan2((P-Q).y, (P-Q).x)); draw(A--B--C--cycle); draw(shift((P+Q)/2)*rotate(angle)*shift(-(P+Q)/2)*ellipse((P+Q)/2,(abs(D-P)+abs(D-Q))/2,sqrt((abs(D-P)+abs(D-Q))**2-abs(P-Q)**2)/2)); draw(P--PA,dotted); draw(P--PB,dotted); draw(P--PC,dotted); draw(Q--QA,dotted); draw(Q--QB,dotted); draw(Q--QC,dotted); draw(P--QA,dotted); draw(Q--PA,dotted); draw(P--QB,dotted); draw(Q--PB,dotted); draw(P--QC,dotted); draw(Q--PC,dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$P_A$",PA,dir(PA)); dot("$Q_A$",QA,dir(QA)); dot("$P_B$",PB,dir(PB)); dot("$Q_B$",QB,dir(QB)); dot("$P_C$",PC,dir(PC)); dot("$Q_C$",QC,dir(QC)); [/asy][/asy] We have the following well known key claim. Claim: Let $\gamma$ be any conic tangent to the three sides of triangle $ABC$. Then, the foci of $\gamma$ are isogonal conjugates with respect to the triangle. Furthermore, given any pair $\{P,Q\}$ of isogonal conjugates, there is a unique conic with foci at $P$ and $Q$ tangent to the three sides of the triangle. Proof: This proof has many cases and directions, so we'll focus on one case to tackle, with the other cases having similar proofs. We'll show that given an ellipse $\gamma$ within $ABC$ tangent to the three sides, its foci $P$ and $Q$ are isogonal conjugates. Let $DEF$ be the tangency points of the ellipse, let $P_A,P_B,P_C$ be the reflections of $P$ over the sides of the triangle, and define $Q_A,Q_B,Q_C$ similarly. We see that $\angle PDB=\angle QDC$, so $PQ_A$ and $QP_A$ intersect at $D$, so $PD+QD=PQ_A=QP_A$, with similar relations holding for the other sides. We see that \[PQ_C^2=AP^2+AQ_C^2-2\cdot AP\cdot AQ_C\cdot\cos\angle PAQ_C=AP^2+AQ^2-2\cdot AP\cdot AQ\cdot\cos\angle PAQ_C,\]and similarly \[PQ_B^2=AP^2+AQ^2-2\cdot AP\cdot AQ\cdot\cos\angle PAQ_B.\]But we have $PQ_C=PF+QF=PE+QE=PQ_B$, so $\cos\angle PAQ_C=\cos\angle PAQ_B$, so $\angle PAQ_C=\angle PAQ_B$. Thus, $AP$ and $AQ$ are isogonal in $\angle BAC$, and by repeating the analysis for the three sides, we learn that $P$ and $Q$ are isogonal conjugates. This completes the proof. $\blacksquare$ The solution to the problem is now trivial as there is a unique conic tangent to the five sides of the pentagon (in the dual space, this is the fact that a conic is determined by five points). Then, the only pair of isogonal conjugates are the foci of this conic.