Does the basis graph of a matroid determine it?

I think this question is answered in the paper A Graphical Representation of Matroids 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).

Let $M$ be the matroid corresponding to the cycle graph $C_3$ with edges $\{a,b,c\}$, so that the bases of $M$ are $\{a,b\}$, $\{a,b\}$, $\{b,c\}$; let $M'$ be the matroid corresponding to the multigraph with two vertices and three edges $\{a,b,c\}$ between them, so that the bases of $M'$ are $\{a\}$, $\{b\}$, $\{c\}$. Then $M$ and $M'$ are not isomorphic (they don't even have the same rank), but they both have $C_3$ as their basis graph. (Note that $M$ and $M'$ are however dual.)