jmc

We have that $$\begin{array}{rl}{z}^2& ={\left(\frac{-\sqrt{3}+i}{2}\right)}^2\\ & =\frac{3-2i\sqrt{3}+i^2}{4}=\frac{3-2i\sqrt{3}-1}{4}\\ & =\frac{2-2i\sqrt{3}}{4}=\frac{1-i\sqrt{3}}{2}.\end{array}$$Then $$\begin{array}{rl}{z}^3& =z\cdot z^2\\ & =\frac{-\sqrt{3}+i}{2}\cdot \frac{1-i\sqrt{3}}{2}\\ & =\frac{-\sqrt{3}+3i+i-i^2\sqrt{3}}{4}\\ & =\frac{-\sqrt{3}+4i+\sqrt{3}}{4}\\ & =i.\end{array}$$Hence, $z^6=i^2=\boxed{-1}.$