Solved with some dihydrogen monoxide Since $AFEO$ and $APNO$ are cyclic, $O$ is the Miquel point of $EFNP$, so $PKOE$ and $FKON$ are cyclic. Now, $\angle EFN=\angle EFO+\angle OFN=90^\circ-\angle ACB+180^\circ-\angle AOM=90^\circ-2\angle ACB+\angle ABC$. Therefore, $\angle EFN+\angle OAC=180^\circ-2\angle ACB$, so the angle bisector of $EF$ and $AO$ is perpendicular to $BC$. This means that if a homothety at $M$ with ratio $2$ maps $I$ to $I'$ and $K$ to $K'$, $AK'\parallel BC$, and if $IM=IO$, then $OI'\parallel BC$, so $AO=I'K'=2IK$. Therefore, it suffices to show $IM=IO$. Let $H$ be the orthocenter of $APN$, and let $D$ be the reflection of $O$ over $I$. Then, since $EDFO$ and $PHNO$ are parallelograms, $\triangle DHO\sim\triangle FNO$. Therefore, $\angle HDO=\angle NFO=\angle ABC-\angle ACB=\angle HAO=\angle HMO$, so $DHMO$ is cyclic. Since $\angle DHO=\angle FNO=90^\circ$, $D$ lies on $BC$, so $I$ lies on the perpendicular bisector of $MO$, which means $IM=IO$.

Very cool problem. It took me a while to realize how $M$ came into the picture. We let $H_A$ denote the reflection of the orthocenter of $\triangle ABC$ over side $BC$ (the intersection of the $A-$altitude with $(ABC)$). We start off with the following minor observations. Claim : Quadrilaterals $ONFK$ and $NKPE$ are cyclic. Proof : This is an easy angle chase. Simply note that, \[\measuredangle OFK = \measuredangle OFE = \measuredangle OAE = \measuredangle OAP = \measuredangle ONP = \measuredangle ONK\]which implies that $ONFK$ is indeed cyclic. The proof that $NKPE$ is cyclic is entirely similar. Now, we are in a position to attack the first part of the problem, via the following key claim. Claim : Point $H_A$ lies on $(AFOE)$ and in particular, $FOH_AE$ is an isosceles trapezoid. Proof : Note that, \[\measuredangle (\overline{OM},\overline{AO}) = \measuredangle AOM = 2\measuredangle ACB + \measuredangle BAC = \measuredangle ABC + \measuredangle ACB = \measuredangle OAH_A = \measuredangle AH_AO\]which implies that $H_A$ also lies on $\omega$ as claimed. Further note that, \[\measuredangle FH_AO = \measuredangle FAO = \measuredangle CAO = \measuredangle H_AAB = \measuredangle H_AAE = \measuredangle H_AFE\]which implies that $OH_A \parallel EF$ and $FOH_AE$ is an isosceles trapezoid as desired. Now, note that since $I$ is the midpoint of $EF$, it lies on the perpendicular bisector of segment $EF$. Since we showed that $EFOH_A$ is an isoceles trapezoid this implies that $I$ also lies on the perpendicular bisector of $OH_A$ so $IO=IH_A$. Now, note that, \[\measuredangle OKE = \measuredangle OPE = \frac{\pi}{2}\]So, $OK \perp EF$. Now, consider the point $R$ such that $OKRH_A$ is a rectangle. $R$ must clearly lie on $\overline{EF}$ as a result of our previous observations. Since $I$ lies on the perpendicular bisector of segment $OH_A$ it must also lie on the perpendicular bisector of segment $KR$. Thus, \[2IK = RK = OH_A=AO\]which finishes the first part of the problem. Next, let $T$ be the intersection of the tangent to $(ABC)$ at $H_A$ and $\overline{BC}$. We first prove the following claim. Claim : Points $E$ and $F$ are respectively the centers of circles $(BH_AT)$ and $(CH_AT)$. Proof : We first note that, \[\measuredangle EH_AB = \measuredangle EH_AO + \measuredangle OH_AB = \measuredangle EAO + \measuredangle OH_AB = \measuredangle BAO + \measuredangle OH_AB = \frac{\pi}{2} + \measuredangle BCA + \measuredangle CBA = \measuredangle H_ABA = \measuredangle H_ABE\]which implies that $EB=EH_A$. Further note that, \[\measuredangle BTH_A = \frac{\pi}{2}+\measuredangle AH_AT = \measuredangle ABC + \measuredangle ACB \]So, $2\measuredangle BTH_A = \measuredangle BEH_A$ which implies that $T$ also lies on the circle through $B$ and $H_A$ with center $E$ proving the claim. The proof that $F$ is the center of circle $(CH_AT)$ is similar. Now, we simply need to piece everything together. Note that, \[TE = EH_A = OF \text{ and } TF = FH_A = OE\]which implies that $TEOF$ is a parallelogram. Thus, $T$ is the reflection of $O$ across $I$, which is simply the $O$-antipodal point in the circle through $H_A$ and $O$ with center $I$. Thus, $T$ lies on this circle as well. To finish off simply notice that, \[\measuredangle MOH_A = \measuredangle ABC + \measuredangle ACB = \measuredangle MTH_A\]so $M$ also lies on $(OH_AT)$ which implies that $IO=IM$ and indeed $\triangle IMO$ is isosceles as required.