TITLE: What is an automorphism group?
QUESTION [13 upvotes]: I'm studying field theory on my own, and I am stuck in the definition of an automorphism and automorphism group. Could you give me examples, like the $Aut(\mathbb Z_7)$? How they are computed?

REPLY [5 votes]: I will first give a very general definition. Let $X$ be an object in a category. Then the automorphism group of $X$ (noted $\mathrm{Aut}(X)$)is the set of invertible morphisms from $X$ to itself. Since we have a composition law, we see that this is indeed a group.
In concrete settings, this has the following interpretations:

If $X$ is a group, then $\mathrm{Aut}(X)$ is the group of all group isomorphisms from $X$ to itself.
If $X$ is a topological space, then $\mathrm{Aut}(X)$ is the group of all homeomorphisms from $X$ to itself. 
If $X$ is a vector space, then $\mathrm{Aut}(X)$ is the group of all linear isomorphisms from $X$ to itself.
If $X$ is a topological group, then $\mathrm{Aut}(X)$ is the group of all continuous group isomorphisms from $X$ to itself.

In some sense, the automorphism group is measuring how much the labelling of the elements in your group (say) matters.

Now, going back to your particular example $\mathrm{Aut}(\mathbb Z_7)$, if you are looking at all field isomorphisms, then as Robert Israel noted, this group is trivial (i.e. $\mathrm{Aut}(\mathbb Z_7)\cong 1$), since $1$ has to go to $1$, and this determines the isomorphism. On the other hand, if you are looking at all group isomorphisms, then remember that $\mathbb Z_7$ is generated (as a group) by $1$, but also by all nonzero residue class. Since you can define a group homomorphism by specifying where a generator goes, you have an isomorphism for each nonzero residue class (namely $1\mapsto p$, where $p\in\{1,\ldots,7\}$). Hence, we have $\mathrm{Aut}(\mathbb Z_7)\cong\mathbb Z_7^\times$.