Does conjugacy persevere being subgroup of each others

Question

If $$K\leq H\leq G$$, is $$K^g\leq H^g$$? where $X^g=\{gxg^{-1}:x\in X\}$ If not, assume $H\triangleleft G$ does the relation hold?

Answer 1

If $x=gkg^{-1} \in K^g$ for some $k \in K$, then $k$ is also in $H$ as $K \leq H$, so that $x=gkg^{-1} \in g H g^{-1}=H^g.$