Determine which pairs of graphs below are isomorphic, if there are any. If two graphs are not isomorphic, explain why using known properties of isomorphic graphs.
I know that two graphs are isomorphic iff they have the same number of cycles, same degree sequence, etc. If I can find a different number of cycles of length $4$ in $J$ than in $G$, then that suffices to show they're not isomorphic. I think there is $1$ cycle of length $4$ in $J$ and there are $2$ cycles of length $4$ in $H$, though I'm not sure how to ascertain this (mathematically, not using a computer algorithm). Clearly, for graph J, if the vertices of a cycle are all consecutive (so that they can be ordered in clockwise order and differ by at most one outermost edge), then the only cycle is (E D C B E). If they're not consecutive, then there should be some reason why they can't form a cycle.
Graph $G$ also seems to have $2$ cycles of length $4$. It has at least $6$ cycles of length $5$: (3 2 1 8 9 3), (4 10 7 6 5 4), (4 10 1 2 3 4), (2 1 8 7 6 2), (2 3 4 5 6 2), (5 6 7 8 9 5). However, I can only seem to find $4$ cycles in $H$ of length $5$: (G F E D C G), (F G H J K F), (G H A B C G), (E F K A B E). Alternatively, one can see that vertex $10$ in graph G is part of at least $3$ cycles of length $5$ and a cycle of length $4$, but there is such vertex in graph $H$, so $10$ cannot be mapped by an isomorphism to an element of $V(H)$ and hence $G$ and $H$ are not isomorphic.
Graph $G$ seems like it's not isomorphic to graph $J$ since there seems to be a different number of cycles of length $4$ in both graphs.
So it seems that $H$ and $J$ are not isomorphic and that $G$ and $H$ are not isomorphic and $G$ and $J$ are not isomorphic.
Is it even correct that no two of the graphs are isomorphic? And if so, would there be a better approach that doesn't involve counting cycles?
The $4$-cycles in the graphs are
So $J$ is immediately not isomorphic to either of the others: too many $4$-cycles.
To prove that $G$ and $H$ are isomorphic, we can start building up an isomorphism. In $G$, vertices $2,6$ are not part of any $4$-cycle; in $H$, vertices $F$ and $G$ are not part of any $4$-cycle. There seems to be enough symmetry that we can just map $2$ to $F$ and $6$ to $G$ (but we might need to backtrack).
Let's try mapping $3$ ($2$'s neighbor) to $E$ ($F$'s neighbor). Then cycle $3,4,5,9$ must be mapped to cycle $B,C,D,E$ somehow; in particular, $5$ must be mapped to $C$, the other vertex on the cycle.
Also, $1$ is now forced to map to $K$ and $7$ to $H$, because those are the only neighbors left of $2$ and $6$ (and $F$ and $G$) respectively.
There is a final arbitrary choice to be made; $4$ must be mapped to a neighbor of $C$ and $E$, which can be either $B$ or $D$, but let's pick $B$. Then $9$ is mapped to $D$, $10$ must be mapped to $B$'s neighbor $A$, and $8$ must be mapped to the remaining vertex $J$.
We can check that mapping $1,2,3,4,5,6,7,8,9,10$ to $K,F,E,B,C,G,H,J,D,A$ respectively is an isomorphism by checking any edges we haven't checked.
As intuition for proving that $G$ and $H$ are not isomorphic, I looked at what happens when we delete edges $26$ and $FG$ in them (these are identifiable graph-theoretically in both as "the only edge between the two vertices not part of any $4$-cycle", so doing this should preserve isomorphism). This leaves two vertices of degree $2$; if we replace those by edges, we get a cube graph in both cases.