Is this graph transformation unique up to isomorphism? [Answer:NO] Suppose, transformation T is defined as given in the diagrams below. 
Every vertex (v) is replaced by deg(v)-gon. And then graph is reconnected as shown.



Those on the left are Gs and on the right are T(G)s.

It is easy to see that every vertex in  T(G) has degree 3. 
This paper claims that graph isomorphism of such graphs can be tested in polynomial time. Also, G can be converted to T(G) in polynomial time. 

Statement I: G1 and G2 are isomorphic iff T(G1) and T(G2) are isomorphic.
EDIT:
Specifications for G1,G2:


G1 = (V1,E1) and G2=(V2,E2) 
|E1| = |E2| and |V1| = |V2|
Sort[{deg(v)|v in V1}] = Sort[{deg(u)| u in V2}]


If Statement I is True then do we have solution for GI problem? 

Note: I am n00b in this field. I invent funny techniques daily.
Along with the already-given answers stating the existence of two graphs G1≠G2 for which T(G1)=T(G2), there is also a different problem: there exist pairs G1=G2 for which T(G1)≠T(G2). This is because the transformation depends not just on the isomorphism class of G, but also on the cyclic ordering of the edges around each vertex of G.

Edited to add an example based on Mark Reitblatt's comment: