Suppose that there is a cyclic group $G$ of order $n$ with a generator $g$. Also suppose that $r$ is a fixed integer. Then define a homomorphism function $f: G \rightarrow G$, $f(x) = x^r$.
How do we get that $f(x) = r \cdot x$?
Quote:
Theorem: Let $G$ be a cyclic group of order $n$ with generator $g$. Fix an integer $r$, and define $f : G \rightarrow G$ by $f(x) = x^r$. This map $f$ is a group homomorphism of $G$ to itself. If $gcd(r, n) = 1$, then $f$ is an isomorphism, and in that case every $y ∈ G$ has a unique $r$th root. More generally, order of kernel of $f = gcd(r, n)$, order of image of $f = n/gcd(r, n)$. If an element $y$ has an $r$th root, then it has exactly $gcd(r, n)$ of them. There are exactly $n/gcd(r, n)$ $r$th powers in $G$. Proof: Since $G$ is abelian the map $f$ is a homomorphism. Use the fact that $G$ is isomorphic to $\mathbb{Z}/n$. Converting to the additive notation for $\mathbb{Z}/n$-with-addition, f is $f(x) = r · x$