TITLE: How to characterize "matching-transitive" regular graphs?
QUESTION [9 upvotes]: I am interested in regular graphs $G$ such that for each pair of 1-factors (=perfect matchings) $F$ and $F'$ there is an automorphism of $G$ that takes $F$ to $F'$. Let's call this property matching transitivity.  
Those graphs should have a fairly big automorphism group. I am wondering whether it is possible to characterize them. 
Trivial examples of matching-transitive graph are $K_{2n}$ or the complete bipartite graph $K_{n,n}$.
EDITED after Brendan McKay's comment:
If we remove successively 1-factors from such a graph, it is clear not true that this property is maintained at each stage.  
The question is inspired by the existence of graphs whose 2-factors are isomorphic (as graphs), i.e. all have the same partition $\pi$ as cycle type. For a discussion and examples of such graphs for various cycle types see here and here. All those graphs are cubic, and BTW I don't think that such a graph can be $k$-regular with $k>3$. (Is there an easy argument for that?)
Some of those graphs have lots of symmetries, others a rather small automorphism group, like for instance, the graph of type $(5,11)$ given in this answer, which has a unique triangle and automorphism group $S_3$, but not less than $12$ different 1-factors (so it cannot be matching-transitive). On the other hand, the Heawood graph ($|Aut(G)|=336$) is, and so is the Coxeter Graph (same automorphism group) as outlined here.
EDIT: As mentioned in another comment, there are many graphs with a unique perfect matching, which is obviously not what I am after. So I'll add the (somewhat mild) condition that each edge should belong to some perfect matching. Such a graph is called $1$-extendable (cf. page 113 in [László Lovász, Michael D. Plummer, Matching Theory, Annals of Discrete Mathematics 29, North-Holland, 1986, ISBN: 0 444 87916 1]).
So: 

For given $n$ and $k$, can anything be said about the matching-transitivity of a connected $1$-extendable $k$-regular graph on $n$ vertices in terms of the size (or structure) of its automorphism group?

Note that both Heawood and Coxeter Graph are indeed $1$-extendable, and so is the $(5,11)$ graph mentioned before. There is possibly more hope now that, roughy speaking, the bigger the automorphism group, the bigger the chance for the graph to be matching-transitive.

REPLY [3 votes]: One infinite family is the lexicographic product (composition) $C_s[\bar K_t]$, where $s$ is odd and $t$ is even. Basically you take the cycle $C_s$ and expand each vertex into a cluster of $t$ independent vertices.  Here is $C_5[\bar K_2]$ with a representative perfect matching. In general, any perfect matching has $t/2$ edges between each consecutive cluster of vertices.  Any two perfect matchings are related by an automorphism that permutes the vertices in each cluster. 

Now I can report on some computations. It turned out to be trickier that expected because of graphs with extremely large groups and matching counts.  For the record I'll give the method.
$G$ is a graph with $n$ vertices and automorphism group $\varGamma$. "Matching" means "perfect matching", and "matching-transitive" means "$\varGamma$ acts transitively on the set of all matchings".
$G$ is matching-transitive iff each of its components is matching-transitive, so let's assume $G$ is connected.
Let $P_1,\ldots,P_k$ be the orbits of the action of $\varGamma$ on the
pairs of edges of $G$.  The "profile" of a matching $M$ is the list $m_1,\ldots,m_k$ where for each $j$ exactly $m_j$ of the $\binom{n/2}{2}$ pairs of edges of $M$ lie in $P_j$. Clearly, every matching in a matching-transitive graph has the same profile (but the converse might not be true).

Compute $\varGamma$.
Start a scan of all matchings, halting when any of these conditions become true: (A) two different profiles have been seen; (B) the number of matchings exceeds $|\varGamma|$; (C) the number of matchings exceeds some huge number chosen to prevent memory overflow; (D) the search completes without (A)-(C) happening.
In the case of (A) or (B), $G$ is not matching-transitive.
In the case of (C), we failed; think of something else to do.
In the case of (D), apply $\varGamma$ to the list of all matchings to test if they are all equivalent.

Case (C) can happen with a matching-transitive graph having a huge number of matchings. I excluded known examples: complete graphs, complete bipartite graphs, $C_s[\bar K_t]$, where $s$ is odd and $t$ is even (defined above). With those cases out of the picture, I have not encountered case (C).
In the case of vertex-transitive graphs, the only examples I found were:

complete graphs, balanced complete bipartite graphs, $C_s[\bar K_t]$, where $s$ is odd and $t$ is even
Petersen graph, Coxeter graph, Heawood graph

These tests did not require that every edge is contained in some matching, but that is true for all the examples.  The graphs tested were:

All transitive graphs to 46 vertices
All cubic transitive graphs to 1280 vertices
All quartic symmetric (arc-transitive) graphs up to 640 vertices

The problem for large graphs is that the my crude backtrack for listing all the matchings starts to take too long. However, all the large cubic or quartic graphs except $C_s[\bar K_2]$ could be eliminated by a different method: use a heuristic to find a hamiltonian cycle and check whether the two (or three, for cubic graphs) matchings implied by the cycle are equivalent.  Sometimes this required testing multiple hamiltonian cycles.
For connected regular graphs that are not vertex-transitive, there are too many trivial cases with very few matchings, so I added the criterion that every edge must be contained in a matching.  For connected non-transitive cubic graphs, here are the counts starting at $n=6$: 0, 1, 1, 2, 3, 4, 8, 8, 14, 24.   For connected non-transitive quartic graphs, the counts starting at $n=8$ are: 0, 1, 2, 6, 5.  I didn't attempt to make any sense out of them.