Let the set $S$ consist of $n(2<n<6)$ complex numbers with modulus 1 satisfying the following two conditions: (1) $1 \in S$; (2) If $z_{1} \in S, z_{2} \in S$, then $z_{1}-2 z_{2} \cos \theta \in S$, where $\theta=\arg \frac{z_{1}}{z_{2}}$. Then the set $S=$
11. $--1,1,-\mathrm{i}, \mathrm{i}\}$.

When $z_{1}=z_{2}=z$, from (2), if $z \in S$, then $z_{1}-2 z_{2} \cos \theta=-z \in S$, since $|z|=1$, so $z \neq -z$. This indicates that $S$ contains an even number of elements. Since $2<n<6$, hence $n=4$.
From (1), we know that $1 \in S$, thus $-1 \in S$.
Let $z_{1}=\cos \alpha+i \sin \alpha(\sin \alpha \neq 0, \alpha \in[0,2 \pi)), z_{2}=1$, then $0=\arg \left(\frac{z_{1}}{z_{2}}\right)=\alpha$. If $z_{1} \in S$, then $z_{1}-2 \cos \theta=\cdots \cos \alpha+i \sin \alpha \in S$, since $\sin \alpha \neq 0$, so $\cos \alpha+i \sin \alpha \neq-(-\cos \alpha+i \sin \alpha)$, hence $\cos \alpha+i \sin \alpha=-\cos \alpha+i \sin \alpha$, which gives $\cos \alpha=0, \sin \alpha= \pm 1$, so $\mathrm{i} \in S,-\mathrm{i} \in S$. Therefore, $S=$ $\{-1,1,-i, i\}$.