I am trying to understand why the conjugacy classes of (continuous) homomorphisms $U(1)\to SU(2)$ are given by the maps $$ \phi_n : z\mapsto \begin{pmatrix} z^n & 0 \\ 0 & z^{-n} \end{pmatrix}, \quad n \in \mathbb Z. $$ Here two homomorphisms $\phi,\psi : G \to H$ of groups are said to be conjugate if $\phi(g) = h^{-1}\psi(g)h$ for some $h\in H$.
Of course, I can see that each $\phi_n$ is a homomorphism, and it is necessary to assume that $n$ is an integer in order for this map to be well-defined and smooth. What I do not understand is why every other homomorphism $U(1) \to SU(2)$ is conjugate to some $\phi_n$.
One idea is to use the spectral theorem to write $$ \phi(z) = h^{-1} D h, $$ where $h\in SU(2)$ and $D = \text{diag}(w,w^{-1})$ for some $w\in U(1)$. Now this would be good enough if $h$ can be chosen to be constant and independent of $z$, since then the problem reduces to finding endomorphisms $z\mapsto w$ of $U(1)$, and I know these are given by $z\mapsto z^n$ (see e.g. $\varphi$ in $\operatorname{Hom}{(S^1, S^1)}$ are of the form $z^n$). But I do not understand how or why $h$ can be chosen independently of $z$ here.
