The statement $xax^{-1} \in H \Longleftrightarrow a \in H$ for all $a \in H$ is logically equivalent to the equality of sets $xHx^{-1} = H$. That might help you understand $K$ better.
For an example where $K \neq G$, let $G = S_3$ (the group of permutations of a set with $3$ elements) and $H = \{e, (1,2)\}$ (the subgroup consisting of the identity permutation and the permutation interchanging $1$ and $2$). To see this, consider the permutation $x = (1,3)$ (interchanging $2$ and $3$). Then $xHx^{-1} = \{e, (2,3)\} \neq H$, so $x \notin K$. As an exercise, you should compute $K$ for this example.