Hm. Like in my comment, I am not sure that my suggestion is the good one, but here's my try :
Take your drawn graph as the first one (which is an amazing picture by the way, good job for putting it there) and define the second graph which is the same but replace $3-2$ by $3-6$ and $6-7$ by $2-7$ (undo the cross, make it parallels). In other words, my new graph would have the edges $ \{0-1, 1-2, \dots, 6-7,7-0, 0-4, 1-5, 3-6, 2-7\}. $
EDIT: Ok here's a proof that they're not isomorphic : your graph has a $5$-cycle ($3-4-5-6-2-3$). My graph doesn't have such a luxury : the sets $A= \{0,2,3,5\}$ and $B=\{1,4,6,7\}$ have the property that every time you travel from a vertice to another using an edge, you must go either from $A$ to $B$ or from $B$ to $A$. Therefore you cannot go back to $A$ in $5$ steps (i.e. a cycle of the form $a-b-c-d-e-f$ will be such that $a \in A \Longleftrightarrow f \in B$, so that $a \neq f$ and this is not a $5$-cycle).
Since this last argument shows that every cycle in my new graph has even length, we could've also used the fact that your graph has a $7$-cycle $(3-2-6-5-1-0-7-3)$.
Hope that helps,