TITLE: Graphs where each edge belongs to the same number of 1-factors
QUESTION [5 upvotes]: Let $G$ be a simple connected graph that has at least one 1-factor. We'll define:
$G$ has property A iff it is edge-transitive.
$G$ has property B iff each edge belongs to the same number of 1-factors.


Is it obvious that $A\implies B$ ?  
Is it possible to characterize the graphs that have B but not A? 


I think that such graphs exist, but haven't been lucky yet. The "natural" candidates like a $C_{2k}$ along with its diagonals of a certain length don't seem to work but some come close to property B.

REPLY [7 votes]: Here is an example for you to enjoy.
Take two disjoint copies of $K_{2,3}$, and form a cubic graph on 10 vertices by joining each of the three vertices of degree 2 in one copy to one of the vertices of degree 2 in the second copy. Here is Sage's picture of this graph.

The graph is not edge-transitive because the three "cross edges" are not equivalent to the 12 other edges.
Each one-factor of this graph must use exactly one of the three cross-edges, e.g. 2-5, and a pair of opposite edges from each of the two four-cycles induced by the vertices on each side other than 2, and 5, making a total of four one-factors using 2-5. 
Similarly there are four using 3-6 and four using 4-7. So we have twelve one-factors, giving a total of 60 edge-occurrences in a one-factor. Twelve of those occurrences were cross edges, leaving 48 occurrences shared between the remaining 12 edges, meaning that each non-cross-edge also appears in four one-factors.