Show that every automorphism of a graph G can be regarded as a permutaion I'm not really good at graphs, and I'm just a beginer at it. There is a problem in book "Graph Theory with Application" By Bondy and Morty says:

Show That every automorphism of a Simple Graph $G$ can be regarded as a Permutation on $V(G)$ which preserves adjacency.
And show the set of these permutations are a group under the usual operation of composition

I tried to do that by making bijections on $G$, but My Solution failed, I just made a subset of all automorphisms.
And I also checked this link and honestly, I couldn't find out the Idea of Coloring:
Example of a simple graph isomorphic to a permutation group.
Can you give me just a little, or maybe a giant, hint?
EDIT:
The definetion of Isomorphism, in here, is:
$G$ and $H$ are isomorph iff there is a bijection $\theta: V(G) \to V(H)$ such that $uv \in E(G)$ iff $\theta(u)\theta(v)\in E(H)$
And an automorphism of $G$ is a isomorphism from $G$ on $G$
 A: 
Show that every automorphism of a simple graph $G$ can be regarded as a permutation on $V(G)$ which preserves adjacency. And show the set of these permutations are a group under the usual operation of composition.

An automorphism maps the vertices of a graph back to the same vertices of the graph. So it is intrinsically a permutation (of vertex labels), and specifically one of the things it must do is preserve adjacency - the only mappings allowed as automorphisms must preserve adjacency, and do that for every vertex across the whole mapping.
Wolfram Mathworld has a useful simple example:

As you can see here the labelling can be accomplished in various ways but always preserves vertex $3$ next to $(1,4,5)$ etc.
So the question is effectively to show that the labelling permutation that maps one of these graphs to another, will always map any of these graphs to another in the set.  Again this seems to be merely a relabelling exercise - just allocate a second label to each vertex that matches the primary configuration and by examining these secondary labels we are already guaranteed that the resulting permutation is an automorphism.
I hope these thoughts help you to clarify your approach to this question.
