What does "Vertex-transitivity" tell us about an arbitrary graph? I know the formal definition, i.e. , for any arbitrary graph G,
$\forall v_i,v_j\in V(G)\, \exists g \in Auto(G) \, s.t. g(v_i)=v_j$
But what does it mean in a broader sense?  
 A: If a graph $X=(V(X),E(X))$ is vertex transitive, then we can talk about some obvious things.


*

*Graph $X$ is regular.(converse is not true Eg: Frucht Graph)

*$Aut(X)$ is non-trivial.

*Graph $X$ is symmetric.

*Graph $X$ is locally similar, that means by looking at the vertices we can actually observe that locally they are same.


For example, Cayley Graph $X=(G,C)$ where $G$ is any group and $C$ is any subset of G which doesn't contain the identity and it's closed under inverse. Cayley graphs are regular and by construction, we can find a subgroup which is acting transitively on $X$.
Another important example is $J(5,2,0)$ Petersen graph. Here $S_5$ acts transitively on the vertex set{ which is nothing but the set of subsets of $\{1,2,3,4,5\}$ of size 2. }.
A: Its very similar to what homogeneity means in  topological spaces.
Basically it means that if I am sitting in one of the vertices and I want to tell Steve which vertex I am at while on the phone, I am gonna be totally screwed, because the graph looks the same from every vertex.
A: Vertex transitive graphs are completely symmetric in the same way as the surface of a sphere is, every node has all the same properties as every other.
Graphs can be used to show group structure, for example:

This shows a dihedral symmetry group. It is the direct/cartesian multiplication of two simpler (sub) groups.
Note that (as a graph) it is also the direct multiplication of two vertex transitive graphs, and so it is also vertex transitive.
Moving the identity element e on the above diagram and relabelling the rest of the graph doesn't change any of the resulting group structure, any of the groups properties/uses, abab is still e, it's all still the same object.
