Yes, the graphs are isomorphic, which you can see by drawing them. One isomorphism is:
This is easy to do if you draw the picture. Here is such a graph:
Any labeling of this graph will be not isomorphic to \(G\text{.}\) For example, we could take \(V'' = \{a,b,c,d,e,f,g\}\) and \(E'' = \{ab, ac, ad, be, cf, dg\} \text{.}\)
The degree sequence for \(G\) is \((3, 3, 2, 1, 1, 1, 1)\text{.}\)
In general this should be possible: the degree sequence does not determine the graph's isomorphism class. However, in this case, I was almost certain this was not possible. That is, until I stumbled up this:
\(G\) is a tree (there are no cycles) and as such also bipartite.
Yes, all trees are planar. You can draw them in the plane without edges crossing.
The chromatic number of \(G\) is 2. It shouldn't be hard to give a 2-coloring (for example, color \(a, d, e, g\) red and \(b, c, f\) blue), but we know that all bipartite graphs have chromatic number 2.
It is clear from the drawing that there is no Euler path, let alone an Euler circuit. Also, since there are more than 2 vertices of odd degree, we know for sure there is no Euler path.