4-subnormal not implies finite-conjugate-join-closed subnormal

Statement
It is possible to have a group $$G$$ and a finite collection of conjugate 4-subnormal subgroups of $$G$$ whose join is not subnormal. In fact, it is possible to have just two conjugate 4-subnormal subgroups whose join is not subnormal.

Facts used

 * 1) uses::3-subnormal not implies finite-automorph-join-closed subnormal (in fact, it is possible to have just two automorphic 3-subnormal subgroups whose join is not subnormal.
 * 2) uses::Left residual of finite-conjugate-join-closed subnormal by normal is finite-automorph-join-closed subnormal

Proof using facts
Suppose that every 4-subnormal subgroup of a group is finite-conjugate-join-closed subnormal. In particular, this means that if $$H$$ is a 3-subnormal subgroup of $$L$$, then whenever $$L$$ is a normal subgroup of a group $$G$$, $$H$$ is finite-conjugate-join-closed subnormal in $$G$$.

Fact (2) says that this implies that $$H$$ is finite-automorph-join-closed subnormal in $$G$$. Thus, we have shown that every 3-subnormal subgroup is finite-automorph-join-closed subnormal. This, however, contradicts fact (1).