Graph Theory question about proving isomorphism without using adjacency matrices Read this question

How do you prove something like this, the only way I can think of at the moment is to write up a complete adjacency matrix and have a program test all permutations to see if there's a match..

It seems my comment answered the question, so here it is as an answer:
To prove two graphs $G_1$ and $G_2$ are isomorphic you need to provide the isomorphism (either explicitly or implicitly), that is, a bijective function $\phi : V(G_1) \to V(G_2)$ such that
$$(v_1,v_2) \in E(G_1) \iff \big(\phi(v_1),\phi(v_2)\big) \in E(G_2).$$
To prove two graphs are not isomophic, you need to show that it is impossible to find such a bijective function.
One approach is to find a distinguishing feature, 
one that cannot be mapped by any isomorphism, 
for example, a vertex of unique degree.
In your case you could consider the number of cycles of length $4$, if the graphs are isomorphic, they have to contain the same number of cycles for any given length (if I didn't make any mistake the middle graph has only $4$ such cycles).
I hope this helps $\ddot\smile$
