$\varphi$ in $\operatorname{Hom}{(S^1, S^1)}$ are of the form $z^n$

Let $f : S^1 \to \mathbb{C}^*$ be a continuous(!) group homomorphism. I claim that $f(z)=z^n$ for some fixed $n \in \mathbb{N}$.

First of all, $f$ corresponds to a continuous group homomorphism $g : \mathbb{R} \to \mathbb{C}^*$ which is constant $1$ on $\mathbb{Z}$ (since $S^1 \cong \mathbb{R}/\mathbb{Z}$), via $g(t)=f\left(e^{2\pi i t}\right)$. There is some $a \in \mathbb{R}$ with $A:=\int_{0}^{a} g(t) dt \neq 0$ (otherwise the derivative $g(a)$ vanishes for all $a$, which is impossible since $g(0)=1$). For every $x \in \mathbb{R}$ it follows $A^{-1} \int_{x}^{x+a} g(t) dt = A^{-1} \int_{0}^{a} g(x) g(t) dt = g(x)$. In particular $g$ is differentiable and satisfies the differential equation $(Dg)(x)=A^{-1} (g(x+a)-g(x)) = A^{-1} (g(a)-1) g(x)$. Thus, there is some $b \in \mathbb{C}$ such that $g(x)=e^{bx}$ for all $x$. Since $1=g(1)=e^{b}$ it follows that $b=2\pi i n$ for some $n \in \mathbb{Z}$, meaning $f(z)=z^n$.

Remark: There are lots of non-continuous group homomorphisms $S^1 \to \mathbb{C}^*$. The reason is that some infinite-dimensional linear algebra and the theory of divisible abelian groups implies that there is an isomorphism of abelian groups $S^1 \cong \mathbb{Q}/\mathbb{Z} \oplus \mathbb{R}^{\oplus \mathbb{R}}$, and there are lots of group automorphisms of $\mathbb{R}^{\oplus \mathbb{R}}$.

If you assume continuity then this follows fairly quickly from the fact that a continuous homomorphism between Lie groups is actually a Lie group homomorphism (i.e. it is automatically smooth). So if $\phi : S^1 \to S^1$ is a continuous group homomorphism then we can consider its differential on the Lie algebra $\phi_* : \mathbb R \to \mathbb R$. Being linear this has to have the form $x \mapsto cx$ for some $c \in \mathbb R$. But by properties of the exponential map of Lie groups, we have $$ \phi(e^{ix}) = e^{i\phi_* x} = e^{icx}. $$ And now for this to be well-defined, we need $c \in \mathbb Z$.

I'm going to post the proof given in the notes posted by Zhen rewritten in my own words:

First note that the proof needs the homomorphisms $S^1 \to S^1$ to be continuous.

(i) If $\alpha : \mathbb{R} \to \mathbb{R}$ is a continuous homomorphism then $\alpha$ is of the form $x \mapsto \lambda x$ for some $\lambda \in \mathbb{R}$. This follows directly from the fact that $\alpha$ is a linear map and one dimensional matrices are multiplication by scalars.

(ii) Continuous homomorphisms $\mathbb{R} \to S^1$ are of the form $e^{i \lambda x}$ for $\lambda \in \mathbb{R}$. To see this note that $(e^{ix}, \mathbb{R})$ is a covering space of $S^1$. Then by the unique lifting property we get that for a continuous homomorphism $f: \mathbb{R} \to S^1$ there is a unique continuous homomorphism $\alpha : \mathbb{R} \to \mathbb{R}$ such that $f = g \circ \alpha$ where $g (x) = e^{ix}$ is the covering map. By (i) we get that $f$ has to be of the form $x \mapsto e^{i\lambda x}$.

(iii) If $\varphi : S^1 \to S^1$ and $\psi : \mathbb{R} \to S^1$ are continuous homomorphisms then so is $\varphi \circ \psi : \mathbb{R} \to S^1$. So we know that $1 = \varphi (\psi (0))$. We also know $\psi$ has to map $0$ to $1$ hence $\psi (0) = e^{i 2 \pi k}$ for some $k \in \mathbb{Z}$. And we also know that $1 = e^{i 2 \pi n}$ for some $n \in \mathbb{Z}$. Hence $\varphi (z) = z^m$ for some $m \in \mathbb{Z}$.