We know that edge colourings of a graph $G$ are vertex colourings of a special graph, namely of the line graph $L(G)$ of $G$. Is there a graph operator $\Phi$ such that vertex colourings of a graph $G$ are edge colourings of the graph $\Phi(G)$ ? I am interested in such a graph operator that can be constructed in polynomial time, i.e. the graph  $\Phi(G)$ can be obtained from $G$ in polynomial time. Remark:  Similar question can be asked for stable sets and matchings. A matching in $G$ is a stable set in $L(G)$. Is there a graph operator $\Psi$ such that stable sets in $G$ are matchings in $\Psi(G)$? Since STABLE SET is $\mathsf{ NP}$-complete and MATCHING belongs to $ \mathsf{P}$, such a graph operator $\Psi$ (if exists) cannot be constructed in polynomial time, assuming $\mathsf{NP}\not=\mathsf{P}$.  EDIT: Inspired by @usul's answer and @Okamoto's and @King's comments, I found a weaker form for my problem: Vertex colourings of a graph $G$ are edge colourings of a hypergraph $\Phi(G)$ defined as follows. The vertex set of $\Phi(G)$ is the samevertex set of $G$. For each vertex $v$ of $G$, the closed neighbourhood $N_G[v]= N_G(v) \cup\{v\}$ is an edge of the hypergraph $\Phi(G)$. Then $G$ is the line graph of the hypergraph $\Phi(G)$ and therefore vertex colourings of $G$ are edge colourings of $\Phi(G)$. Again, I am grateful for all answers and comments showing that, with or without assuming $\mathsf{NP}\not=\mathsf{P}$, the operator I am looking for cannot exist. It would be nice if I could accept all the answers!
Are vertex colourings--in a sense--edge colourings?
graph theory;graph algorithms
By analogy with the line graph, I think you are asking the following:For every undirected graph $G = (V,E)$, does there exist an undirected graph $G' = (V',E')$ such that each vertex $v \in V$ corresponds to an edge $(v_1,v_2) \in E'$ and the edges corresponding to $u \in V$ and $v \in V$ share at least one endpoint if and only if $(u,v) \in E$?The answer can be seen to be no. Consider the four-vertex tree $G$ with root $v$ having three children $x,y,z$. In $G'$, we must have four edges: $(v_1,v_2),(x_1,x_2),(y_1,y_2),(z_1,z_2)$. Further, it must be the case that either $v_1$ or $v_2$ is an endpoint of each of the other three edges (i.e., $\left| \{v_1,v_2\} \cap \{x_1,x_2\} \right| \geq 1$, etc). But this means that at least two of the other three edges must share a common endpoint, which violates our requirements since no two of $x,y,z$ are adjacent in the original graph.I think the same graph will give you a counterexample for the matching question as well.