What is the "order" of an automorphism of a group?

Can anyone please tell me what is the "order" of an automorphism of a group? I skim through some books but can't find the clear definition for this notion.

Thanks so much and I really appreciate if you give me some material about automorphism...

If $G$ is a group and $g \in G$, then the order of $g$, denoted $|g|$, is the smallest positive integer $n$ such that $g^n = e$, where $e$ is the identity.

The collection of automorphism of a group $G$, denoted $\text{Aut}(G)$, is a group under function composition. The order $\sigma \in \text{Aut}(G)$ is $|\sigma|$, i.e. order of $\sigma$ as an element of the group $\text{Aut}(G)$.

The order of a group is the cardinality of its underlying set. In the case of an automorphism group, it is the cardinality of the set of all automorphisms. I.E. (finitely many automorphisms) the number of isomorphisms from a particular group to its self. There is a special kind of automorphism (self isomorphism) called an inner automorphism. An automorphism is an inner automorphism iff it's conjugation by a group element. As it turns out, the set of inner automorphisms is a normal subgroup of the set of automorphisms, and, as a group, is isomorphic to $\frac{G}{Z}$, where Z is the center of the group. The center of a group is the group of elements that commute with everything.

Exercises:

1. Prove that Inn(G) (group of inner automorphisms) is normal in Aut(G). (just grind it out)

2. Prove that the center of a group is a (normal) subgroup of G

3. Prove that Inn(G) is isomorphic to G mod the center by considering a the natural map: $g\rightarrow \chi_g$, where $\chi_g$ is conjugation by g. (prove it's a homomorphism, prove that it's onto (trivial), find its kernel, use the 1st isomorphism theorem.

Bonus: (If you don't know the 1st isomorphism theorem) Let G, H be groups and $\phi: G\rightarrow H$ be a surjective homomorphism. Prove that H is isomorphic to G mod the kernel.