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Let $G$ be a group. The set of all automorphisms of $G = \operatorname{Aut}(G$), with $(\operatorname{Aut}(G), \circ)$ 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 if $f:C_n\rightarrow C_n$ is a homomorphism of $C_n$ then $f=f_m$ for some $m$.

Hint: show that if $f(g)=g^k$ then $f=f_k$.

I do not quite know what to do. I know when $r=1$, $f_m(g)=g^m$ but can't see if this helps.

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    To know the image of $g$ i enough since it generates $C_n$ – 2017-02-22

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Recall that a homomorphims is such that $f(gh)=f(g)f(h)$. Thus by induction we have that $f(g^r)=f(g)^r$.

Now suppose $f$ is an homomorphism from $C_n\to C_n$. Since $C_n=\langle g\rangle$ we then have $f(g)= g^k$ for some $0\leq k \leq n$. Therefore $f(g^r)=f(g)^r=(g^k)^r= g^{rk}$ and $f=f_k$.

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    I know that $f(gh)=f(g)f(h)$ but I can't see exactly how you got $f(g^r)=f(g)^r$ – 2017-02-22
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    The base case for induction is $f(g)=f(g)$. Now assume for some $k\in \Bbb N$, $f(g^k)= f(g)^k$. Thus we have that $f(g^{k+1})=f(g g^k)=f(g)f(g)^k=f(g)^{k+1}$. Hence by induction you have the result. – 2017-02-22
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    What is your $f(h)$ in $f(g^r)=f(g)^r$? – 2017-02-22
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    In my proof that that formula holds you mean? It would in that case be $g^k$. – 2017-02-22
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    I understand from the induction _why_ $f(g^r)=f(g)^r$ but I can't see how you decided to use the fact $f(g^r)=f(g)^r$ in the first place. Where did it come from? Just your own intuition? – 2017-02-22