Actually the first direction is easy. I think the answer is:

We show that f.A.f^(-1)= B. i.e., for every g in A, f.A.f^-1 is in B ( i.e. f.A.f^-1 fixes L ) and for every h in B, f.B.f^-1 is in A (i.e. fixes E).

Doing the first part, let x be an element in L then f^-1(x)= y in E , and g fixes y , and f again send y to L, so f.A.f^-1 fixes L and similarly for the other assertion, hence the first direction is proved . I am having trouble in the reverse direction.