I want to show that any $X\in SU(2)$ is conjugate to a matrix of the \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix} for $\theta\in \mathbb{R}$.
So I guess I want to find $K\in SU(2)$ st. $X=K\begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}K^{-1}$. I've tried to diagonalize $X$ and taking the $\exp$ function, but it doesn't seem to get me anywhere. Any hint is appriciated.
If $X\in SU(2)$, then its characteristic polynomial is a quadratic polynomial. Therefore, it has some root $\lambda\in\mathbb C$. Let $v$ be a unit vector that is an eigenvector of $X$ with eigenvalue $\lambda$. Let $w$ be an unit vector orthogonal to $v$. Since $X$ is unitary, $w$ is also an eigenvector of $X$; let $\mu$ be its eigenvalue. Then$$1=\det X=\lambda\mu.$$So, $\lambda\neq0$ and $\mu=\frac1\lambda$. Furthermore, since $X$ is unitary, $v$ and $Xv(=\lambda v)$ have the same norm (which is $1$). So, $\lvert\lambda\rvert=1$, which means that $\lambda=e^{i\theta}$ for some $\theta\in\mathbb R$. So, $\mu=\frac1\lambda=\frac1{e^{i\theta}}=e^{-i\theta}$. Can you take it from here?