Let $M$ be a matroid with set of basis $\mathcal{B}$. The *basis graph* of $M$ is a graph with set of vertices $\mathcal{B}$ and edges $(B,B')$ always that $B$ and $B'$ differ (as sets) by exactly one element.


It is easy to see that this graph can be thought as the $1$-skeleton of the basis polytope of $M$. 


My question is if it is indeed true that the isomorphism class of $M$ is determined by its basis graph. To avoid trivialities, we ask $M$ to be loopless and bridgeless (it is $M^\*$ is also loopless), because otherwise there are simple counterexamples.


My intuition says the answer is no, but I couldn't find a counterexample so far.


EDIT: As pointed out in the answers and comments, since duality preserves the basis graph, it is reasonable to ask if there is a pair of indecomposable non isomorphic non dual matroids (loopless and bridgeless) such that they induce isomorphic basis graphs.



Does the basis graph of a matroid determine it?
I think this question is answered in the paper [A Graphical Representation of Matroids](https://doi.org/10.1137/0125060) by C. A. Holzmann, P. G. Norton, and M. D. Tobey. From the abstract:


"A base graph of a matroid is the graph whose points are the bases of the matroid. Two bases are adjacent if they differ by exactly one element. A definition of equivalence of matroids is given and it is shown that two matroids are equivalent if and only if their base graphs are isomorphic. In particular, if M and $M\_1 $ are nonseparable matroids with isomorphic base graphs, then M is isomorphic to either $M\_1 $ or its dual..."


The notion of equivalence is $M = \sum\_i M\_i$ and $M' = \sum\_i M'\_i$ are equivalent if and only if $M'\_i$ is isomorphic to $M\_i$ or its dual for each $i$ (and some reordering if needed).


