I am trying to figure out a lower bound on the number of different bases a matroid can have. The matroid has rank $n$ and its ground set is the disjoint union of two bases.
I think that it's $2^n$ because we can take any element from the first basis and exchange it with another element from the second basis (since the two bases are disjoint and have the same cardinality n, any element in the second basis would fit in the first - following from the basis exchange property).
Yes. An example of a matroid that achieves this lower bound is $U_{n,n}$ where every element has another element placed in parallel.
To see that you can't go below this lower bound, if your ground set is partitioned into bases $B_0$ and $B_n$, you can perform basis exchange $B_{i} = (B_{i-1} - e_i) \cup f_i$, where $e_i \in B_0$ and $f_i \in B_n$. Having $n$ binary choices of whether to use $e_i$ or $f_i$ in a basis gives you $2^n$. If there are fewer than $2^n$ bases, that means that there is some set $\{e_1, e_2, \ldots, e_j\} \subseteq B_0$ that is inadmissible as a sequence for the exchanges described above, and that contradicts the basis exchange axioms.