Finding order of $Aut(\mathbb{Z}_{3})$ with just the very basics

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In my Abstract Algebra class, our professor just introduced the concepts of groups to us, and one of the problems on our homework assignment is to compute the order of the automorphism group $Aut(\mathbb{Z}_{3})$ of the additive group $\mathbb{Z}_{3}$. I've been searching online and in books how to do this, because our professor did not tell us. What I have found is that most of the time this is done by applying one of many theorems, including using the fact that $\mathbb{Z}_{3}$ is cyclic (We haven't learned what a cyclic group is yet.), that it's isomorphic to $U_{3}$ (We haven't learned yet how that even helps).

Basically, it looks like I have to do this by using the proverbial "stone knives and bear skins". I know that $\mathbb{Z}_{3}$ is an additive group, consisting of the elements $\{0, 1, 2\}$, and that an automorphism is a bijective map from a group to itself that preserves the group structure. So, I suppose the identity map that would map $\mathbb{Z}_{3}$ to itself would work. But, how do I list out what all the possible bijective homomorphisms from $\mathbf{\mathbb{Z}_{3}}$ to itself without using any slick tools? There must be a lot of them!

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What do you know about the image of $0$ under a group homomorphism?

Do you know how many bijective maps exist on a $3$-element set?

Can you combine this knowledge to narrow down the possibilities? Do you know how to check those possibilities?

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Lemma. Let $G$ and $G'$ be two groups. Let $\varphi:G\rightarrow G'$ be a group morphism. One has: $$\varphi(1_G)=1_{G'}.$$

Proof. One has: $$\varphi(1_G)=\varphi(1_G\times 1_G)=\varphi(1_G)^2.$$ Right-composing the equality by $\varphi(1_G)^{-1}$ leads to the result. $\Box$

Therefore, an element of $\textrm{Aut}(\mathbb{Z}_3)$ must send $\overline{0}$ to $\overline{0}$. Then, you are free to send $\overline{1}$ on $\overline{1}$ or $\overline{2}$, otherwise the map won't be bijective. Afterward, check that the two constructed maps are morphisms.