Show that a graph cannot be a Cayley graph

Question

How can I show that the following graph cannot be a Cayley graph:

I think it's not a cayley graph because it is not vertex transitive, I just don't know how to show it (or explain it more formally).

Answer 1

*This has to come as a comment to the previous answer,I don't have enough reputation to make a comment to answers.

Please look at the definition of vertex-transitivity, it says for any two vertices say $v_{1},v_2\text{ }\exists \text{ } f\in Aut(X)$ such that $f(v_1)=v_2$.An automorphism is an isomorphism from a graph to itself, so it has to preserve subgraph structures.So suppose vertex $v_1$ is incident to three triangles and $v_{2}$ is incident to 2 triangles , we can't map $v_1$ to $v_{2}$ since it wont be preserving subgraph structures hence it's not vertex-transitive.

There is another way to prove this, since your graph has $|V(X)=7|$,so $V(X)=\mathbb{Z_{7}}$ and it's a degree 4 connected graph,you have to choose a $C,$with $|C|=4$ and $<C>=\mathbb{Z_7}$.For any $C$ which satisfies the above condition we will fail to find an isomorphism between $X(G,C)$ and this graph hence graph cannot be cayley.

Answer 2

Hint: One way to show it is not vertex-transitive is to count how many triangles different vertices are in.