 Solution

A one-dimensional continuous group representation is a continuous group homomorphism $\rho :S^1\to {\mathbb{C}}^{\times }$. Its image is compact. If some $|\rho (z)|$ differed from one, its positive or negative powers would be unbounded, so the image lies in $S^1$. Lift $t\mapsto \rho (e^{it})$ to a continuous real argument $a(t)$ with $a(0)=0$. Then $a(s+t)-a(s)-a(t)$ is a continuous integer multiple of $2\pi$ and vanishes at $(0,0)$, hence is always zero. The continuous Cauchy functional equation gives $a(t)=ct$. Periodicity forces $e^{2\pi ic}=1$, so $c=n\in \mathbb{Z}$. Consequently $\boxed{\rho (z)=z^n}$; conversely every such map is a continuous one-dimensional group representation.