Automorph-conjugacy is not intersection-closed

Statement with symbols
We can have a group $$G$$ with two automorph-conjugate subgroups $$H, K \le G$$, such that $$H \cap K$$ is not automorph-conjugate in $$G$$.

Example in the symmetric group
(This example demonstrates the stronger fact that automorph-conjugacy is not conjugate-intersection-closed).

Let $$G$$ be the symmetric group on six letters: $$\{ 1,2,3,4,5,6 \}$$. Let $$H, K$$ be the following 2-Sylow subgroups of $$G$$:

$$H = \langle (1,3,2,4), (1,2), (5,6) \rangle ;\qquad K = \langle (1,2), (3,5,4,6), (5,6) \rangle$$

In other words, $$H$$ is the internal direct product of a 2-Sylow subgroup on $$\{ 1,2,3,4\}$$ with the 2-Sylow subgroup on $$\{ 5,6,\}$$, while $$K$$ is the internal direct product of the 2-Sylow subgroup on $$\{ 1,2 \}$$ with a 2-Sylow subgroup on $$\{ 3,4,5,6 \}$$.

The intersection is given by:

$$H \cap K = \{ (1,2), (3,4), (5,6) \}$$.

Now, note that:


 * Both $$H$$ and $$K$$ are automorph-conjugate, because they are both Sylow subgroups, and Sylow implies automorph-conjugate.
 * $$H \cap K$$ is not automorph-conjugate. To see this, note that $$G$$ has an outer automorphism that sends transpositions to triple transpositions. Under this automorphism, $$H \cap K$$ goes to a subgroup of $$G$$ that contains three commuting triple transpositions. If this is conjugate to $$H \cap K$$, then $$H \cap K$$ should also contain three commuting triple transpositions. But it doesn't.