Let $G = \langle g \rangle$ be a cyclic group of order $n$. Suppose $m$ is an integer such that $\gcd(m, n) = 1$. Define $\phi: G \to G, g^i \mapsto g^{mi}$. Then $\phi$ is (I think) an automorphism. What can we say about the order of $\phi$?
My attempt:
Note that $\phi^x(g^i) = g^{m^x i}$. The exponentiation $m^x$ brings to mind Euler's Theorem, which is applicable in this case since $m, n$ are coprime: $m^{\Phi(n)} \equiv 1 \bmod n$, where $\Phi$ is Euler's totient function. So $\phi^{\Phi(n)} \equiv I$, where $I$ is the identity function, and $\Phi(n)$ is a candidate for the order of $\phi$. However, trying out some simple examples shows that we can have a number $x$ less than $\Phi(n)$ such that $\phi^x \equiv I$, so at least in some cases $\Phi(n)$ isn't the order of $\phi$. I'm not sure what else to try now. Any suggestions/corrections would be appreciated.
Best Answer
First, since OP's comment stated that the kernel is trivial, it remains to prove that $\phi$ is surjective to show that it is in fact an automorphism. Let $g^j\in G$. We want to show that there exists some $g^i\in G$ such that $\phi(g^i)=g^j$, that is $g^{mi}=g^j$. This means that $mi\equiv j\bmod n$. Since $m$ and $n$ are coprime, there always exists a solution for $i$ in the equation. This shows that $\phi$ is surjective.
In fact, let $\psi$ be a homomorphism on $G$ such that $\psi$ maps $g$ to $g^m$, it can be proven that $\psi$ is an automorphism of $G$ iff $m$ is coprime to $n$. This is because if $\psi$ is an automorphism, then $\langle \psi(g)\rangle=G$, that is, $\langle g^m\rangle=G$. Therefore, $m$ must be coprime to $n$.
This means that every automorphism of $G$ has the property you stated and this implies that $|\text{Aut}(G)|=\Phi(n)$. Hence what we can only conclude for the order of $\phi$ in Aut$(G)$ is that by Lagrange's Theorem, the order of $\phi$ divides $\Phi(n)$.