TITLE: Removal of non-isomorphic edges results in the same graph
QUESTION [24 upvotes]: There exists a (simple unlabeled) graph on 6 nodes with a pair of non-isomorphic edges (i.e., there is no graph automorphism that sends one edge into the other) such that removal of either of them results in the same graph:

The two non-isomorphic edges here are colored red and blue. This is the smallest example of such graph and it is unique for 6 nodes.
I wonder what would be the smallest example of a graph with a triple of pairwise non-isomorphic edges and removal of any one of them resulting in the same graph.
P.S. See also sequence https://oeis.org/A245246 -- extension is welcome.
UPDATE: user2097 below gave a construction for a graph on $(\tbinom{n}{2}+1)\cdot n$ nodes that squares the number of such pairwise non-isomorphic edges. When applied for the aforementioned graph on 6 nodes, it gives a graph on 96 nodes with a quadruple of pairwise non-isomorphic edges, removal of each of which results in the same graph. But it is probably not the smallest such graph.

REPLY [7 votes]: This is essentially a comment on user2097's excellent answer. One consequence of that answer was the construction of a graph with four non-isomorphic edges whose removals result in isomorphic graphs. The graph so constructed had 96 nodes and 293 edges. 
Without changing the idea of their construction, various tricks allow an improvement to a graph with the same property with 24 nodes and 47 edges.
The tricks used are:

Instead of connecting the nodes $i$ to nodes $v_{ec}$ whenever $i$ is in $e$, do this whenever $i$ is in $e$ and $c$ is in some set of vertices stable under the isomorphism of subgraphs.
Instead of replacing all edges with a copy of $G$ or $G'$, only do this for enough edges in the complete graph to contain the two special edges and be stable under the isomorphism of subgraphs.

To wit, begin with the graph described by the original poster:

Replace the solid green edges with a copy of $G$ as follows:

Replace the dotted green edge with a copy of $G$ without one of its special edges:

The final result is as promised.

It is not hard to see by starting at the 1-valent vertices that this graph has no automorphisms.