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.