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Let $C_n$ be a cyclic group of order $n$, I want to find all isomorphism from $C_n$ to $C_n$.

If $f: C_n\rightarrow C_n$ is an isomorphism, then for all $x\in C_n$, it exists a generator $g$ s.h $\:f(x)=f(g^m)=f(g)^m,\:m\in \mathbb{N}$

Now because $f(g)\in C_n$, in order to generate the whole group it must be true that $\text{gcd}(n,m)=1$ so they can be $\phi(n)$ automorphism. More specific the automorphism are $f_1(x)=x,\ldots, f_{\phi(n)}(x)=x^{n-1}$

Now I want to show that the set of those automorphisms, form a cyclic group with order $\phi(n)$ and operation the composition.

$f_1(x)=x$ is the identity element and the inverse of $f_i(x)$ is (don't know)

also I don't know how to prove that $\text{Aut}(C_n)$ is cyclic, don't see why some $\langle f^k\rangle$, $k\in \mathbb{N}$ generates all the other isomorphisms

Arturo Magidin
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    I note that each automorphism is a permutation of the elements of $C_n$. Is this group of permutations cyclic? Is there a choice of $m$ such that $g \mapsto g^m$, $g \mapsto g^{2m}$, $g \mapsto g^{3m}$, ... is the group of automorphisms? I note that not every cyclic group's subgroup of units has a primitive root. – Eric Towers Nov 13 '21 at 23:50
  • @EricTowers $m=1$? But only primes have primitive roots, I still don't quite follow – 領域展開 Nov 13 '21 at 23:54
  • "But only primes have primitive roots". False. $2$, $4$, powers of odd primes and twice powers of odd primes have primitive roots. – Eric Towers Nov 13 '21 at 23:55
  • @EricTowers oh ok i see, how primitive roots are relative here?. Every comment here is a new riddle. I need to sleep sometime, thanks for trying to help thought. – 領域展開 Nov 13 '21 at 23:59
  • It is not true that the automorphism group of a cyclic group is necessarily cyclic. So what you want to prove is not true, which will make it rather difficult to prove. – Arturo Magidin Nov 14 '21 at 00:13
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    A previous question of yours establishes that $\mathrm{Aut}(C_n)$ is isomorphic to $\mathbb{Z}_n^\times$. However, the latter is rarely cyclic. When is it cyclic? See here: https://math.stackexchange.com/questions/314846/for-what-n-is-u-n-cyclic – Randall Nov 14 '21 at 00:42
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    Also, that same question answers this one. There, you saw that the autos were the homomorphisms $f:C_n \to C_n$ where $f(g)=g^m$, where $m$ has a multiplicative inverse mod $n$ (and $g$ is a generator that is fixed before the whole argument begins). That's it: that's the description of all the automorphisms of $C_n$. – Randall Nov 14 '21 at 00:53

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