What does Aut$(\Bbb Z)$ look like? What does Aut$(\Bbb Z)$ look like? (Integers with the operation of addition)
I understand that it's the set of all automorphisms from $\Bbb Z$ to $\Bbb Z$, or Aut$(\Bbb Z) = \{\alpha_1, \alpha_2, ... : \alpha_i$ is an isomorphism from  $\Bbb Z$ to $\Bbb Z \}$.
I figured that the only isomorphisms that work are $\alpha_1(n) = n$ or $\alpha_2(n) = -n$ because those are the only bijective operation preserving mappings.  Then Aut$(\Bbb Z) = \{n, -n\} = \{a_1, a_2\}?$
 A: If ${\bf Z}$ is viewed here as an additive group, then this group is cyclic.  Therefore, any automorphism of the group must map a generator to a generator.
The group has only two generators: $1$ and $-1$.  Therefore, there is only one nontrivial automorphism: the one mapping $1$ to $-1$.
A: You are trying to determine all automorphisms (isomorphisms) of the group $\mathbb{Z}$. So say that $\phi: \mathbb{Z} \to \mathbb{Z}$ is an isomorphism. You want to figure out what $\phi$ could be.
Now, say that $n = \phi(1)$. Then $\phi(m) = \phi(1+\dots + 1) = m\phi(1) = mn$, so $\phi$ is completely determined by its value at $1$ and you see that the image of $\phi$ is exactly all multiples of $n$. For $\phi$ to be surjective (onto), we need $n$ to be $1$ or $-1$. Hence you only have the two isomorphisms that I think you listed in your question. That is
$$
\phi_1(1) = 1 \\
\phi_2(1) = -1.
$$ 
Now $Aut(\mathbb{Z})$ is then a group (under function composition) with two elements. You might know that there is (up to isomorphism) only one group of order $2$, namely $\mathbb{Z}_2$. Hence
$$
Aut(\mathbb{Z}) \simeq \mathbb{Z}_2.
$$
