The uniform matroid $U(n,k)$ has $n$ elements and all $k$-element subsets are bases. I was under the impression that $U(2k,k)$ is self-dual, that is, it is isomorphic to its dual in which bases are replaced by their complement.
However, the usual definition of matroid isomorphism seems to be the following:
Definition. Two matroid $(X_1,I_1)$ and $(X_2,I_2)$ are isomorpic, if there exists a bijection $X_1\to X_2$ that induces a bijection $I_1\to I_2$.
It is not hard to see, that this defintion cannot provide an isomorphism between, e.g. $U(4,2)$ and its dual. No permutation of the elements of the matroid can induce the following map on the characteristic vectors of the bases:
$$0011\longleftrightarrow 1100,\qquad 1010\longleftrightarrow 0101,\qquad 1001\longleftrightarrow 0110.$$
What now? Is $U(4,2)$ not self-dual? Or is there a different version of matroid isomorphism, e.g. the following:
Definition. Two matroids $(X_1,I_1)$ and $(X_2,I_2)$ are isomorphic if their lattices of independent sets are isomorphic as lattices.
And if so, which one is the "standard definition"?
I think you are hoping that the map that does the dualizing already is the isomorphism if a matroid is self dual. That doesn't need to be the case. $U(2k,k)$ and its dual are isomorphic through the identity map but that is a different map than the one that maps each element in $U(2k,k)$ to its dual.