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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 $r\in \mathbb{Z}, f_m(g^r):=g^{rm}$

List the elements of Aut$(C_4)$ and Aut$(C_{12})$.

I know I can use the fact that $f_m$ is an automorphism of $C_n$ if and only if gcd$(m,n)=1$.

So for Aut$(C_4)$ $n=4$ and $m=3$, $g^4=1$, and $g^3 $ has order 4? This is as much as I can do, I don't really know where to go from here.

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You pretty much have it. You did miss $f_1$ though.

The $(C_12)$ case follows similarly, just list all $f_I$ with $\gcd (12, I)=1$.

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    So the elements of $Aut(C_4)$ are $\{1,3,4\}$? and the elements of $Aut(C_{12})$ are $\{1,5,7,11\}$? How do we know its order? It doesn't seem that these groups are cyclic? – 2017-02-21
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    $4$ is not an element of $Aut(C_4)$. The order of the group is the Euler phi function of $n$. – 2017-02-21
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    Sorry my mistake. I am a bit confused because I am asked to classify these groups up to isomorphism and do not know how. – 2017-02-21
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    Well if you are stuck, there are only so many groups of order $2$ and $4$. You can always just check to see which ones they are. If after that you are still stuck, google the Klein 4 group. – 2017-02-21