B

作$C_{1}D\bot OA$于$D$，则$\angle C_{1}DO=90^{\circ}$，

$\because$四边形$OABC$是菱形，$O\left(0,0\right)$，$A\left(-4,0\right)$，菱形绕点$O$顺时针旋转$90^{\circ}$后得到菱形$OA_{1}B_{1}C_{1}$，
$\therefore \angle AOC=\angle A_{1}OC_{1}=\angle B=60^{\circ}$，$OC=OC_{1}=OA=4$，
$\therefore \angle C_{1}OD=30^{\circ}$，
$\therefore C_{1}D=\frac{1}{2}OC_{1}=2$，$OD=\sqrt{3}C_{1}D=2\sqrt{3}$，
$\therefore$点$C_{1}$的坐标为$(2\sqrt{3}$，$2)$.
故选：$B$.