In the page 128, of the book Analysis on Lie groups by James Faurat, we have the following paragraph:
Every element $x\in SU(2)$ is conjugate to a diagonal matrix of the form
$a(\theta) = \exp(\theta X_{1})$ = $\begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{i\theta} \end{pmatrix}$
that is, $x = ga(\theta)g^{-1}$ with $g\in SU(2)$, $\theta\in\mathbb{R}$. In fact a unitary matrix $x$ is normal: $x^{*}x = xx^{*}$, hence diagonalisable in an ortogonal basis. This means that
$x$ = $g \begin{pmatrix} e^{i\theta_{1}} & 0 \\ 0 & e^{i\theta_{2}} \end{pmatrix} g^{-1}$, $\,\,\,\,(g\in U(2),\,\, \theta_{1}, \theta_{2}\in\mathbb{R})$
One can choose $g$ with determinant equal to one and, since $\det(x) =1$, one can choose $\theta_{2} = -\theta_{1}$:
$x$ = $g \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} g^{-1}$, $\,\,\,\,(g\in SU(2),\,\, \theta\in\mathbb{R})$
Question:
Why can we choose $g\in SU(2)$?
Because every unitary matrix is normal and the spectral theorem says that if $M$ is normal, then there is an unitary matrix $U$ such that $U^{-1}MU$ is diagonal. Now, let $\Delta=\det U$ and let $\omega$ be a square root of $\Delta$. Then $\frac1\omega U$ will also work and $\frac1\omega U\in SU(2)$.
This works for $n\times n$ complex matrices (take a $n$th root of $\Delta$ in this case).