Symmetric group of degree six or higher is not weak subset-conjugacy-closed in general linear group over rationals

Statement
Suppose $$n \ge 6$$. Consider the symmetric group $$S_n$$ of degree $$n$$ as a subgroup of the general linear group of degree $$n$$ over the field of rational numbers, denoted $$GL(n,\mathbb{Q})$$, via the embedding as permutation matrices. Then, $$S_n$$ is not a weak subset-conjugacy-closed subgroup of $$GL(n,\mathbb{Q})$$. In other words, we can find subsets $$A$$ and $$B$$ of $$S_n$$ that are conjugate as subsets in $$GL(n,\mathbb{Q})$$ but not in $$S_n$$.

We can take $$A$$ and $$B$$ to be subgroups.

Similar facts

 * Analogue of Brauer's permutation lemma fails over rationals for every non-cyclic finite group: In fact, the statement generates a very large collection of examples, though for our purposes we need only one example.

Opposite facts

 * Brauer's permutation lemma guarantees that any two subgroups of $$S_n$$ that are conjugate in $$GL(n,\mathbb{Q})$$ have an isomorphism between them that preserves cycle types of elements.

Example for $$n = 6$$
Consider the following two subsets of the symmetric group $$S_6$$ acting on the set $$\{ 1,2,3,4,5,6 \}$$:

$$A = \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$$

and

$$B = \{, (1,2)(3,4), (1,2)(5,6), (3,4)(5,6) \}$$

These are not conjugate inside $$S_6$$, because $$A$$ has two global fixed points while $$B$$ has none. However, they are conjugate inside $$GL(6,\mathbb{Q})$$. We can see this either by finding explicit matrices that perform this conjugation, or by noting that the two representations have the same character, and hence, since the Klein four-group is a rational representation group, they must be conjugate over the rational numbers.

Example for higher $$n$$
We can use the same example as for $$n = 6$$, fixing all the $$n - 6$$ elements.