Determine if a set of graphs are isomorphic 
I am trying to use adjacency matrices to show if the graphs are isomorphic. I have created the matrices but am finding it hard to interpret them to find a pair (or more) of isomorphic graphs for each A, B and C. I am comparing the matrices to see if they can be "rearranged" to look like another matrix. Is this a poor way to check if graphs are isomorphic?? 
Isn't graphs that are isomorphic able to be "redrawn" to look like each other? Would this make all graphs in A and B not possible to be isomorphic?
 A: The easiest way to show two graphs aren't isomorphic is to find a property of the graph that one has and the other doesn't. For example, consider the first row of graphs. The first graph has a chain of vertices in a line that have, in order, degree $1$, then $2$, then $3$. Neither of the other two graphs have this, so the first graph is different from the second two. The second graph has a vertex of degree three, while last graph doesn't have any vertices of degree $3$, so they are non-isomorphic as well.
To show that two graphs are isomorphic, you need to find an isomorphism between them. You're correct that in both the first two sets of graphs, all three graphs are non-isomorphic.
A: Your assessment of parts (a) and (b) is correct. It sounds like you need some additional tools to describe the differences.
One tool is degree sequence, which simply lists the various degrees of the vertices in sorted order. If two graphs have a different degree sequence, they are not isomorphic. However two graphs that have the same degree sequence may or may not be isomorphic.
The only example here is the right-hand graph in part (a) has a different degree sequence to its counterparts, since it has no degree-3 vertex.
You can also check which vertex degrees are adjacent. So there is vertex of degree 3 in the left and centre graphs of part (a), but their adjacent vertices have degrees (2,2,1) and (2,1,1) respectively, so they are not isomorphic.
You can use similar tools to identify the graphs of part (b) as non-isomorphic.
In part (c), the graphs are all cubic (every vertex is degree 3) so you will need to think about the cycles that the vertices are part of, in particular the shortest cycle for each vertex. The right-hand graph is the famous Petersen graph.
At this point you may need to produce an explicit isomorphism. This consists of labelling the vertices and producing a mapping such that the corresponding vertices are connected in both graphs.
A: To prove two graphs are not isomorphic, we can:

*

*Choose a vertex invariant, i.e., a property of the vertices of graphs which remains unchanged by relabelling the vertices;
Figuring out which to choose is a bit of an art (and trial and error), but since the vertex degree, i.e., number of neighbors, seems ineffective in the later graphs in this case, maybe something slightly more complicated (e.g. the number of vertices at distance two) would be better.


*Compute the vertex invariants for each vertex in each graph; and


*If they have different multisets of entry invariants, they're not isomorphic.
To prove two graphs are isomorphic, we need to identify an isomorphism.  To do this, we:

*

*For one graph, $G$ say, we label the vertices distinctly $1,2,\ldots,n$;


*For the other graph, $H$ say, we label them using $1,2,\ldots,n$, ensuring that vertices labelled $i$ and $j$ are neighbors if and only if they are neighbors in $G$;


*The isomorphism is then "the vertex labeled $i$ in $G$ maps to the vertex labeled $i$ in $H$".
