Finite normal subgroups of $\operatorname{SU}(n)$
This question is already answered by Qiaochu Yuan in your second link. Any finite normal subgroup of $\textrm{SU}(n)$ must be contained in its center. Since $Z(\textrm{SU}(n))\cong \mathbb{Z}/n\mathbb{Z},$ these are exactly the subgroups of $\mathbb{Z}/n\mathbb{Z},$ that is the groups of matrices generated by $\omega_m I$ for $m|n$ where $\omega_m$ is a primitive $m$th root of unity.