Let $\sigma_1,\sigma_2,\sigma_3$ be the Pauli matrices. Prove that $U\sigma_2 U^\top=\sigma_2$ for all $U\in\operatorname{SU}(2)$.
I can prove that $U\sigma_2 U^\top= \pm \sigma_2$ using the fact that $U\sigma_2 U^\top$ is skew-symmetric and that the determinant is $1$, but I can't differentiate between $+1,-1$.
Let $u,v\in\Bbb C$ such that $|u|^2+|v|^2=1$. Then $$U\sigma_2U^\top=\begin{pmatrix}u&v\\-\bar v&\bar u\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}u&-\bar v\\v&\bar u\end{pmatrix}=\begin{pmatrix}i(-uv+uv)&-i(u\bar u+v\bar v)\\i(u\bar u+v\bar v)&i(\bar u\bar v-\bar u\bar v)\end{pmatrix}=\sigma_2.$$