Graph transformation related to graph isomorphism Basically got graph transformation related to graph
isomorphism.
Define $G \to G'$.  $V(G')=V(G) \cup E(G)=\{v_1\ldots v_n\} \cup \{e_1\ldots e_m\}$. Call $v_i$ vertices $v'$ and $e_i$ vertices $e'$.
Edges of $G'$.
(1) Add $(v_i,e_j)$ iff $v_i \in e_j$.
This graph is bipartite and is the subdivision of $G$. According to a paper
this preserves GI.
(2) Make a clique of $v'$ vertices, i.e.
add $(v_i,v_j)$ for $i \ne j$ and without
multiple edges. This graph is chordal and
split and according to paper preserves GI.
(3) Make a clique of $e'$ vertices, i.e.
add $(e_i,e_j)$ for $i \ne j$ and without
multiple edges.
Vertices of $G'$ can be partitioned
into two cliques on $v',e'$ and
the edges between the cliques are from (1),
are the subdivision of $G$.
$G \cong H \implies G' \cong H'$.

Q1 Does this transformation preserves isomorphism? $G' \cong H' \implies G \cong H$?

I am inclined to believe so, since the cliques doesn't matter
much and the subdivision subgraph of (1) preserves isomorphism.
A graph is $X$-free if it doesn't contain induced subgraph
$X$.
Claim 1. $G'$ is $(P_4 \cup K_1)$-free.
$3$ or more $v'$ vertices induce triangle and $3$ or more $e'$ vertices
induce triangle since they are in a clique.
The $5$ vertices of $(P_4 \cup K_1)$ can't be partitioned
in $v',e'$ without a triangle.

Is this correct?

Conjecture based on experimental data and suggested proof:
If $G$ is triangle-free, $G'$ is $\overline{3K_2}$-free.
$\overline{3K_2}=K_{2,2,2}$ and the clique number is $3$,
so if induced $\overline{3K_2}$ would exists it must be on $3e' \cup 3v'$
vertices. I suspect that the $v'$ vertices must induce triangle in $G$
for the following reason: The partitions of $K_{2,2,2}=\overline{3K_2}$ are
$(v_1,e_1),(v_2,e_2),(v_3,e_3)$. The edges $(v_i,v_j),(e_i,e_j)$
come from the cliques. The remaining edges are 
$(v_1,\{e_2,e_3\}),(v_2,\{e_1,e_3\}),(v_3,\{e_1,e_2\})$,
which is $C_6$, subdivision of triangle, contradicting triangle-free.
Counterexamples are welcome.
For triangle-free $G$, $G'$ is $(P_4 \cup K_1,\overline{3K2})$-free.
Appears to me correctness of the above and
paper top of p. 10
would imply GI is in P.
 A: Here is a proof that the answer to Q1 is yes.  
Let $f$ be the described transformation and suppose that $G'=f(G)$ for some $G$.  Let $A$ and $B$ be the vertices of $G'$ that correspond to the vertices and edges of $G$ respectively.  Note that $(A,B)$ is a partition of $V(G')$ such that $A$ is a clique, $B$ is a clique and each vertex in $B$ has exactly two neighbours in $A$.  Call such a partition good. 
Claim. $G'$ has a unique good partition unless $G$ is a star or $2$-regular.    
Proof. Let $(A',B')$ be a good partition different from $(A,B)$.  Write $A'=A_1 \cup B_1$ and $B'=A_2 \cup B_2$ where $A_1 \cup A_2=A$ and $B_1 \cup B_2=B$.  First suppose that for some $i$, both $A_i$ and $B_i$ are non-empty.  Since, $A_i \cup B_i$ is a clique, it follows that $B_i$ are the edges of a star, and $A_i=\{x\}$ where $x$ is the middle vertex of this star.  Since the claim obviously holds if $|V(G)| \leq 2$, the only remaining possibilities are 


*

*$(A',B')=(B,A)$, or

*$|A_1|=1$ and $B_2=\emptyset$, or

*$|A_2|=1$ and $B_1=\emptyset$.


The third possibility is impossible since the vertices in $B_2$ would only have one neighbour in $A'$.  The first possibility implies that $G$ is $2$-regular, and the second possibility implies that $G$ is a star. This completes the proof of the claim.
Note the above proof shows that if $G$ is $2$-regular or a star, then $G'$ has exactly two good partitions. For a good partition $(A,B)$ let $G'(A,B)$ be the graph obtained by removing the edges of the cliques on $A$ and $B$.  If $G'$ has two good partitions $(A,B)$ and $(A',B')$, it is easy to see (and somewhat magical) that $G'(A,B) \cong G'(A',B')$.  Therefore, we can recover $G$ from $G'$ by finding a good partition $(A,B)$ and computing $G'(A,B)$ (this will be the graph obtained from $G$ by subdividing every edge once).    
