Need to understand the statement: "Symmetry group of all complete graphs is vertex transitive." I am reading John Meier's "Groups, Graphs and Trees" book to brush up my basics in Combinatorial Group theory. I am confused with the definition of the symmetry group of graphs as well as the above-mentioned statement.
Due to lack of proper vocabulary and understanding, I am stuck in the definition 1.17. 
All I understand is that SYM(G) is the group of permutations of edges and vertices of a graph G and  let H be a subgroup of SYM(G), then H is vertex-transitive if there exist an automorphism $\alpha$ s.t $\alpha(v) = v^{'}$.
Exactly my confusion is regarding the visualisation of such automorphism in  an actual graph such as this complete graph. What exactly permuation looks like visually in a graph? and then I want to organically figure out what definition 1.17 is trying to tell.
 A: I don't have access to the book your are mentioning, so I'll follow the definitions on wikipedia on graph automorphisms. I'm using simple, non-directed graphs here, where all the concepts are relatively easy.
So a graph automorphism of the graph $(V,E)$ is a permutation $\pi$ of the vertex set $V$, such that
$$\forall u,v \in V: (u,v) \in E \Longleftrightarrow (\pi(i),\pi(v)) \in E.$$
So it's a permutation of the vertices such that for any pair $(u,v)$ of vertices, the permutation preserves the status of "there is/isn't an edge between $u$ and $v$".
There exist equivalent definitions, where you permutate the edge set $E$ a well (call that permutation $\sigma$) and require that if the edge $e$ is between vertices $u$ and $v$ then $\sigma(e)$ must be between $\pi(u)$ and $\pi(v)$.
It's equivalent for simple, non-directed graphs, but for graphs that can have multiple edges between the same pair of vertices you need to consider the edge permutations as well. From your formulation, it seems the book works with the more general formulation that includes edge permutations.
How does that apply to the graph you used as an example? I included some vertex numbering for clarity:

First, not that this is not a complete graph! A complete graph has an edge between each pair of vertices, but this one hasn't (the edge between veritces $1$ and $2$ is missing, among others).
From the definition of a graph automorphism, it should be clear that image $\pi(v)$ of a vertex $v$ has the same degree as the orignal vertex $v$. That graph has 3 vertices with degree $1$ ($1,2$ and $3$) and one with degree $3$ (vertex $4$). So any graph automorphism needs to map vertex $4$ to itself.
OTOH, you can permutate the vertices $1$ to $3$ among themselves freely, they are all connected to vertex $4$ and only to that vertex, so any permutation will not change that.
So the automorphism group (or another name: symmetry group) of that graph consists of the 6 permutations of the $4$-element vertex set $\{1,2,3,4\}$ that have $4$ as a fixed point.
So this group is not vertex transitive: vertex $1$ can not be mapped to vertex $4$, for example, and vertex $4$ can not be mapped to vertex $3$. There are more maps that can't happen, but even one is enough to show that it's not vertex-transitve. But since the graph is not complete, that is not a contradiction to the statement in your title.
Now how can one verify that the statement in your title is correct? The easiest way to see that is by noticing that for the complete graph the symmetry group of the graph is the symmetric group $S_V$ of the vertex set. That means any permutation of the vertext set is also an automorphism of the graph.
That should be clear because the answer to the question "Given a pair of vertices, is there an edge between them?" is always "Yes"! So in the initial definition of a graph isomorphism, the part $ (u,v) \in E$ is always true, and so is $(\pi(i),\pi(v)) \in E$. So they are equivalent, as required.
In the symmetric group, you can of course always find a permuation (more precisly $(|V|-1)!$ many) that maps a given vertex $v$ to another (or the same) vertex $v'$.
