AC
The line $y=-\sqrt{3}(x-1)$ intersects the $x$-axis at $(1,0)$, so the focus of the parabola is at $(1,0)$, thus $p=2$. Option A is correct. Since $k_{MN}=-\sqrt{3}$, the inclination angle of line $MN$ is $120°$, so $|MN|=\frac{2p}{\sin^2 120°}=\frac{16}{3}$. Option B is incorrect. Draw a perpendicular from $M$ to the directrix $l$, meeting $l$ at $M'$. Draw a perpendicular from $N$ to the directrix $l$, meeting $l$ at $N'$. Let $P$ be the midpoint of $MN$ and draw a perpendicular from $P$ to the directrix $l$, meeting $l$ at $P'$. Connect $MP'$ and $NP'$. By the definition of a parabola, $MF=MM'$ and $NF=NN'$, so $|MN|=|MM'|+|NN'|$. By the midline of a trapezoid, $PP'=\frac{1}{2}(|MM'|+|NN'|)=\frac{1}{2}|MN|$, so $PP'=MP=PN$. Thus the circle with $MN$ as diameter is tangent to $l$, and C is correct. From the diagram, $\triangle OMN$ is clearly not isosceles, so D is incorrect.