Show that any automorphism of $\mathbb{Z}_n$ is equal to multiplication by $f(1)$ mod $n$.

Question

In other words, I'm trying to prove that $f(x) = f(1)\cdot_nx$ for all $x$ in $\mathbb{Z}_n$ and $f\in Aut(\mathbb{Z}_n)$. Here, $\cdot_n$ denotes multiplication mod $n$.

I'm completely stuck. I've done several examples to try to understand what this is even saying, but I'm not even sure if my examples are right. Consider $\mathbb{Z}_4$ and $x = 3$. Then $f(3) = (f(1)\cdot 3)\mod 3$. Since $f$ is an automorphism, we know it maps the identity element to the identity element, so $f(1) = 1$ I think. So $f(3) = 3 \mod 3 = 0$, but this makes no sense to me. I thought there were only 2 automorphisms in $\mathbb{Z}_4$, which are the identity and $(1 3)$.

EDIT: Just realized I should have written 3 mod 4, which is 3. So $f(3) = 3$. This is just the identity mapping, though, right?

Basically, I don't even know where to start.

Answer 1

Here is an attempt to understand what may be your doubt,and clear it.

So we will consider $\mathbb Z_4$. The question is saying the following : for all $x$, $f(x) = x \cdot f(1) \mod n$.

Which means, it does not make sense to fix a value of $x$.

What can vary in your question? Well, $f(1)$ can vary, right? And it can take values between $1$ and $4$, since $Z_4 = \{[1],[2],[3],[4]\}$.

Now, we will fix values of $f(1)$ and see the resulting automorphism.

$f(1)=1$: This gives $f(x)=x$, or the identity.

$f(1)=2$: This is not an automorphism, since $f(2)=f(4)=0$, so the kernel is non-zero.

$f(1)=3$: This maps $1 \to 3$, $2 \to 2$, $3 \to 1$ and $4 \to 4$.

$f(1)=4$: This is the zero map, it's left to you to check. Obviously not an automorphism.

Hence, we have got the two automorphisms ,identity and $(13)$ from this setup. I hope now you have understood what the requirements of the question are.

As for the question itself, it is simply a question of generation, so I will leave you with a hint. Since $[x] = x \cdot [1]$ for all equivalence classes $[x] \in Z_n$, the place where $[1]$ maps decides the place where the rest of the $[x]$ map, since automorphisms have the property $f([x]) = f(x \cdot[1]) = x \cdot f([1])$. Now you can formalize why any automorphism must be of the form given in the question.

Answer 2

Let $f:\Bbb Z_n\to \Bbb Z_n$ is an automorphism .Then $f([0]=[0]$ since homomorphisms and hence automorphisms map identity to identity.

$\Bbb Z_n$ is cyclic and generated by $[1]$.

So $f([a])=f(a[1])=af([1])(\mod n)$

where homomorphisms vary as $f([1])$ varies ;some giving automorphisms and some not.