Realized graph of majority of permutations

I don't have a full proof, but I wand to share some thoughts and ideas to start a discussion.

Firstly note that each directed acyclic graph corresponds to a permutation, which can be found as a topological ordering of this graph.

Also set of edges of a simple digraph can be divided into two subsets inducing an acyclic graph each. For this purpose choose any vertex ordering along a line, then collect arcs going from left to right into one subset, and arcs going from right to left into another one (let's call it left-right separation). Therefore as long as digraph $G$ is simple the permutations $(1, 2, \ldots, n)$ and $(n, n - 1, \ldots, 1)$ induce its supergraph $H$.

So a potential problem comes with double arcs only. Here we need to divide set of single arcs into two subsets such that each of two subsets being united with the set of double arcs induce an acyclic graph. Here we can't take an arbitrary ordering of vertices to use left-right separation as before.

An example is the following: permutations $(1, 2, 3)$ and $(3, 1, 2)$ induce a graph $G$ with arc multiset $\{\,(1, 2), (1, 2), (1, 3), (2, 3), (3, 1), (3, 2)\,\}$. And a bad ordering for left-right arc separation is $1, 3, 2$: two subsets are $\{\,(1, 3), (3, 2)\,\}$ and $\{\,(2, 3), (3, 1)\,\}$. The first subset perfectly unites with the arc $(1, 2)$. But the second one produces a cycle $1, 2, 3, 1$. However two other vertex orderings, $1, 2, 3$ and $2, 1, 3$, are good for left-right separation, and remaining three are essentially the same as these three.

This example shows that it makes sense to require that vertices of the same weak component of graph $G_{\mathrm D}$, induced by double arcs of $G$, form a contiguous subsequence of vertex ordering. But it is not enough: consider permutations $(1, 2, 3, 4)$ and $(1, 4, 2, 3)$. This example is essentially the same as the previous one, but all vertices now belong to the same weak component in $G_{\mathrm D}$.

Here I give another approach. It could be wrong, feel free to challenge it. Let's construct a vertex ordering in the following way. For every position from $1$ to $n$ we select a vertex with no incoming double arc going from a remaining vertex. Among all such vertices we give priority to a vertex with maximum number of incoming double arcs from previously selected vertices. After getting such an ordering let's do left-right separation of single arcs. One subset being united with double arcs set definitely induces an acyclic graph, since all double arcs go from left to right. And I suppose that another one being united with double arcs set also induces an acyclic graph, but a proof is not ready at this moment.

An important property of the graph $G_{\mathrm D}$ is that it is a transitive closure of an acyclic digraph. The property about transitive closure comes from an observation that if $x$ comes before $y$ in all permutations and $y$ comes always before $z$ than $x$ always comes before $z$. And existence of a cycle leads to a contradiction of form "$x$ always comes before itself".

I believe the result is true. Consider first what sort of graphs are realised by collections of 2 permutations. For a permutation $\sigma$, let $E_\sigma = \{(\sigma(i),\sigma(j)): 1 \leq i < j \leq n \}$ be the "corresponding edge set". Then one way of interpreting the multigraph we form from a collection of 2 permutations $\sigma$ and $\sigma'$ is that it is the graph with edge set $E_\sigma \sqcup E_\sigma'$. Consequentially, for any pair of vertices, either $x \Rightarrow y$, $y \Rightarrow x$ or $x \leftrightarrow y$ (I hope the meaning behind this notation is clear here!)

Thus, our strategy is simple. For any collection of permutations which realises a graph G, it is sufficient to pick our 2 permutations to realise $H$ such that whenever $x \Rightarrow y$ in $G$, $x \Rightarrow y$ in $H$, and otherwise, $x \leftrightarrow y$. Then $H$'s edge set covers that of $G$.

To construct the two permutations, label the vertices $\{v^i_{j}: 1 \leq j \leq n_i\}$ (where $i$ ranges from $1$ to some $k$ and $n_1 + \dots + n_k = n$) such that if we look at the subgraph of G comprising of edges which appear twice (which we note is a forest), the sets $\{v^i_{j}: 1 \leq j \leq n_i\}$ are the trees (possibly singletons) that make up the forest. Then, with a bit of thought, we can realise that the required permutations can be chosen to be $\sigma = (v^1_1,\dots,v^1_{n_1},\dots, v^k_{1},\dots, v^k_{n_k})$ and $\sigma' =(v^k_1,\dots,v^k_{n_k},\dots, v^1_{1},\dots, v^1_{n_1})$ (we reverse the order of the trees). It is not hard to see that vertices within the same tree appear as double edges in the realised graph, while vertices not in the same tree have edges both ways in the realised graph.

Apologies if I've misunderstood something! Just let me know.