How to describe automorphism groups? How should I Describe automorphism groups $\operatorname{Aut}(\mathbb{Z}_9)$ and $\operatorname{Aut}(\mathbb{Z}_3 \times \mathbb{Z}_3)$ ? 
If  a group G contains a normal subgroup H of order 9, G is generated by H and an element x not in H of order 3, How should I classify all such groups G?
 A: Ok, let's take the two separately.
Begin by noting that $\text{Aut}(\mathbb{Z}_9)\to \mathbb{Z}_n^\times$ (where the latter is the unit group) given by $f\mapsto f(1)$ is an isomorphism. So, you know that $\text{Aut}(\mathbb{Z}_9)\cong \mathbb{Z}_9^\times$. Now, it is a common fact the invertible elements of $\mathbb{Z}_n$ are those coprime to $n$. Thus, you at least know that $|\mathbb{Z}_9^\times|=\varphi(9)=6$. Since there is only one abelian group of order $6$ you may conclude that $\mathbb{Z}_9^\times\cong\mathbb{Z}_6$. 
Now, the case of $\text{Aut}(\mathbb{Z}_3^2)$ is a little bit tricker. The key is to notice that $\mathbb{Z}_3^2$ is just a two-dimensional $\mathbb{Z}_3$-vector space and that group maps are really just $\mathbb{Z}_3$-linear maps. So, the invertible group maps $\mathbb{Z}_3^2\to\mathbb{Z}_3^2$ are precisely the invertible $\mathbb{Z}_3$-linear maps $\mathbb{Z}_3^2\to\mathbb{Z}_3^2$. But, this last set of maps has a nice description. Namely, by picking a basis for $\mathbb{Z}_3^2$, say $\{(1,0),(0,1)\}$ you can realize isomorphically the set of $\mathbb{Z}_3$-linear maps as just the $2\times2$ invertible matrices over $\mathbb{Z}_3$. Thus, you see that, up to isomorphism, $\text{Aut}(\mathbb{Z}_3^2)$ is just $\text{GL}_2(\mathbb{Z}_3)$.
Now, both of these results have generalizations. The second one is easier. Namely, using the exact same logic, I bet you can prove that for any prime $p$ and any integer $n$ that $\text{Aut}(\mathbb{Z}_p^n)\cong\text{GL}_n(\mathbb{Z}_p)$. 
The first is a little harder. You can actually prove, using the exact same technique I mentioned, that $\text{Aut}(\mathbb{Z}_n)$ ($n$ some integer) is just $(\mathbb{Z}_n)^\times$. Now, to find $\mathbb{Z}_n^\times$ it suffices to find $\mathbb{Z}_{p^m}^\times$ for $p$ a prime. This is because, as rings, $\mathbb{Z}_n$ splits into a product of rings of the form $\mathbb{Z}_{p^m}$ (by the Chinese Remainder Theorem). Since this is a ring isomorphism, the unit groups of the two are isomorphic and since the unit group of a product is the product group of the units you get that $\mathbb{Z}_n^\times$ is just a product of groups of the form $\mathbb{Z}_{p^m}^\times$. Now, to find these is a little bit tricky, but you do get the following result:
$$\mathbb{Z}_{p^m}^\times\cong\begin{cases}\mathbb{Z}_{p^{m-1}(p-1)} & \mbox{if}\quad p\text{ is odd}\\ \{e\} & \mbox{if}\quad p=2, m=1\\ \mathbb{Z}_2 & \mbox{if}\quad p=2,m=2\\ \mathbb{Z}_{2^{m-2}}\times\mathbb{Z}_2 & \mbox{if}\quad p=2,m\geqslant 3 \end{cases}$$
If you are really interested in the proof, you can see my blog post about it here. 
