This is easily complex-bashable, I assume a synthetic sol along the lines of the complex bash is possible but I'm too lazy to find it. Let $A_1,B_1,C_1$ be the midpoints of $OA',OB',OC'$ respectively. Then $A_1,B_1,C_1$ are on the circle with diameter $OI$, so the circumcenter of $\triangle A'B'C'$ is $I$. Let $A=x^2,B=y^2,C=z^2$ on the unit circle such that $\overline{AI}\cap (ABC)=-yz$ and similar relations hold. $A'=x^2-yz,B'=y^2-xz,C'=z^2-xy,I=-xy-yz-xz$. The centroid of $\triangle A'B'C'$ is $\frac{x^2+y^2+z^2-xy-xz-yz}{3}$, so the nine-point center of $\triangle A'B'C'$ is $\frac{x^2+y^2+z^2}{2}$, which is the nine-point center of $\triangle ABC$.