Having trouble understanding the logical structure of this statement 
Let $H$ be a subgroup of $G$, and let $K = \{x \in G: xax^{-1} \in H \iff a \in H\}$.

Does this mean that $K$ is all of $G$?  The statement seems to say nothing about $x$ itself.  If there is no $a \in H$ that $x$ conjugates, then the antecedent is false so that statement is trivially true.
 A: 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.
A: "If there is no $a \in H$ that $x$ conjugates" makes no sense. You can use $x$ to conjugate any element $a$; the question is what the result will be, and the statement says that $K$ contains the elements $x$ of $G$ such that conjugating $a$ with $x$ yields an element of $H$ iff $a$ is in $H$.
A: $$xax^{-1} \in H \iff a \in H$$
means that if $a$ is in $H$, then $xax^{-1}$ is in $H$, and if $a$ is an element of $G$ such that $xax^{-1}$ is in $H$, then $a$ is also in $H$.  So $x$ is in $K$ if the above happens.  To contrast, $x$ is not in $K$ if there exists $a\in H$ such that $xax^{-1}$ is not in $H$, or if there exists $a\in G\setminus H$ such that $xax^{-1}$ is in $H$.  
