It would be useful if someone could go through my proof and tell me if this is adequate and if not how could I improve. There a similar proofs on here but none that show both ways for an iff proof.
Let $G$ be a group. The set of all automorphisms of $G$ = Aut($G$), with (Aut($G), \circ$) also being a group.
Consider $C_n=\langle g:g^n=1\rangle$, the cyclic group of order $n$. For each positive integer $m$ with $1\leq m\leq n$ define a map $f_m:C_n\rightarrow C_n$ as follows: for $r\in \mathbb{Z}, f_m(g^r):=g^{rm}$
Show that $f_m$ is an automorphism of $C_n$ if and only if gcd$(m,n)=1$.
My attempt: So for the first way round we suppose $f_m$ is an automorphism and deduce that gcd($m,n)=1$
Since we suppose $f_m$ is an automorphism, then injectivity holds. Injectivity holds if and only iff the kernel is trivial. Take the generating element $g$. Let $y=g^r$, an arbitary element with $0\leq r \leq n-1$.
Suppose $y^m=e$, then $y^m=g^{rm}=e$, with $n|mr$, since the kernel is trivial then $r$ must $= 0$, thus gcd$(m,n)=1$. (don't know if I use $r$ or something else since $r$ is in the question)
Now we check the other way, where we suppose that gcd$(m,n)=1$ and deduce that $f_m$ is an automorphism: If gcd$(m,n)=1$, then this implies the kernel is trivial (don't know if I can say this without more proof first?)
For finite sets of the same cardinality, an injective function is automatically surjective which gives us bijectivity and part i) shows $f_m$ is a homomorphism. Therefore the properties for an isomrphism from $C_n$ to itself are met and $f_m$ is an automorphism of $C_n$.