This problem looks very funny and cool and I wonder if it is true for any other trio of triangle centers? $\textbf{Solution}$ So $M$ is $I-$Dumpty of $\triangle{HIO}$ Let $E$ midpoint $OH$. $N$ is the reflection of $I$ over $M$. Easy to see $HION$ is harmonic $\implies \triangle{IEH} \sim \triangle{ION}$ Let $X$ midpoint $IH$ so $\triangle{IEX} \sim \triangle{IOM}$. $\implies \frac{IO}{OM}=\frac{IE}{EX}=\frac{IE}{\frac{1}{2}IO}$ $\implies OM = \frac{IO^2}{2IE}=\frac{R^2-2Rr}{R-2r}=R$. Hence $M$ lie on $(ABC)$. $Q.E.D.$

Attachments:

For isogonal pairs $P,P’$ and $Q,Q’$ center of spiral similarity taking $PQ$ to $Q’P’$ lies on $(ABC)$.

This is a rather well known problem.

What is the point P for triangle ABC?

Upon inversion at $I$ we may check by angles that $O,H$ are sent to antigonal conjugates, and $I$ becomes the orthocenter of $ABC.$ In particular the midpoint of the images of $O,H$ must be on the nine-point circle. Now $M$ forms a parallelogram with $I$ and the images of $O,H$ so it is the reflection of $I$ over a point on the nine-point circle, thus it is on the circumcircle.

Let $IM\cap HO =T$. We know that $\angle{HIM}=\angle{IOM}=\alpha$ and $\angle{MIO}=\angle{IHM}=\theta$. By $\triangle{HMI}\sim \triangle {IMO}$ we know $|MO||MH|=|IM|^2$. By $\angle{HMI}=\angle{IMO}$ with bisector teo we know $|MT|^2=|MO||MH|-|HT||TO|$. After these we got $|HT||TO|=(|IM|+|MT|)(|IM|-|MT|)$. Let the symettry of $I$ to $M$ ve $L$. The equality we fond Last moment says us $HIOL$ is cylic. Let $\angle{OHI}=x_1$ and $\angle{HOI}=x_2$. Its clear $\angle{MOL}=x_2$ and $\angle{MHL}=x_1$. $\angle{IHO}=\angle{LHM}=x_1$ so $HT$ is symeddian in triangle $IHL$. So $\frac{|IT|}{|TL|}=\frac{|HI|^2}{|HL|^2}$. Similarly $\frac{|IT|}{|TL|}=\frac{|IO|^2}{|OL|^2}$. So $\frac{|HI|^2}{|HL|^2}=\frac{|IO|^2}{|OL|^2}$. With angles we can see that $\triangle {MHL}\sim \triangle {MLO}$. So $\frac{|OL|}{|HL|}=\frac{|ML|}{|HM}=\frac{|MO|}{|ML|}$ and we get $\frac{|OL|^2}{|HL|^2}=\frac{|MO|}{|HM|}$ and $\frac{|IO|^2}{|HI|^2}=\frac{|MO|}{|HM|}$. With bisector teo in $\triangle {HMO}$, $\frac{|MO|}{|HM|}=\frac{|OT|}{|TH|}$ we found so $\frac{|IO|^2}{|HI|^2}=\frac{|OT|}{|TH|}$. Thats why $IT$ is symeddian in $\triangle {HIO}$. Let $R$ be the middle of $[OH]$. $\angle{RIO}=\alpha$ and $\angle{HRI}=\theta$ by symeddian. Let the intersection of line that goes on $R$ and parallel to $IH$ and $IO$ be $Q$. By $\triangle {IRQ}\sim \triangle {OMI}$ similarity we can see that $\frac{|OI|^2}{2|IR|}=|OM|$. By feurbach and eulers teo. we know $|IR|=\frac{R}{2}-r$ and $|OI|=\sqrt{R^2-2Rr}$. If we put them in the equation we get $|OM|=R$. We are done.

We have $MHI\sim MIO$ since $\measuredangle IHM=\measuredangle OIM$ and $\measuredangle MOI=\measuredangle MIH$. Work on the complex plane. Denote by $j$ the complex coordinate of $I$. $a=x^2,b=y^2,c=z^2,j=-yz-zx-xy,o=0,h=x^2+y^2+z^2$. \[\frac{m-h}{m-j}=\frac{m-j}{m}\iff \frac{m-x^2-y^2-z^2}{m+xy+yz+zx}=\frac{m+xy+yz+zx}{m}\iff m^2-m\sum{x^2}=m^2+2m\sum{xy}+(\sum{xy})^2\]\[\iff m(\sum{x})^2=-(\sum{xy})^2\iff m=-\frac{(\sum{xy})^2}{(\sum{x})^2}\]\[\overline{m}=-\frac{\sum{\frac{1}{x^2y^2}+2\sum{\frac{1}{x^2yz}}}}{\sum{\frac{1}{x^2}}+2\sum{\frac{1}{xy}}}=\frac{\sum{x^2}+2\sum{xy}}{\sum{x^2y^2}+2\sum{x^2yz}}=\frac{1}{m}\]As desired.$\blacksquare$