Solved with porkemon2. Redefine $D$ as the point on $BC$ with $MD \perp EF$. First suppose that $EF$ is the perpendicular bisector of $MD$. Then if lines $EF$ and $BC$ intersect at $T$ we have $TM = TD$ so $\angle MTF= \angle FTD$. Moreover, since $M$ is the Miquel point of complete quadrilateral $BEFC$ quadrilateral $CFMT$ is cyclic so the angle condition implies $CF = FM$. Then $\angle ACM = \frac{1}{2} \angle AFM$ so $\angle AFM = \angle AOM$ and $O$ lies on $\odot(AFM)$ as required. If we are given that $O$ lies on $\odot(AEF)$ then $2\angle ACM = \angle AOM = \angle AFM$ giving $CF = FM$ and we can reverse the steps above.

2015 China TST 1.1