Let $H$ be a subgroup of $G$. Let $K = \{x \in G: xax^{-1} \in H \iff a \in H\}$. Prove that $K$ is a subgroup of $G$. Closure under multiplication:

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Let $a, x, y \in H.$  Since $H$ is closed under inverses and multiplication, $xax^{-1} \in H$ and $yay^{-1} \in H.$ Since $H$ is closed under multiplication, $xyax^{-1}y^{-1} \in H.$ 

$\leftarrow$

Suppose $xyax^{-1}y^{-1} \in H$, then $a \in H$ since $H$ is closed under multiplication. So, $xy \in K.$

Closure under inverses:

Since $H$ is closed under multiplication and inverses, if $a \in H,$ then $x^{-1}ax \in H.$ If $x^{-1}ax \in H,$ then $a \in H$ because $H$ is a group. So, $x^{-1} \in K.$

Identity:

Suppose $a \in H.$ Then $a = eae^{-1} \in H.$ So, $e \in H.$

I'd like to see how I can fix/improve this proof.
You need to show just two things: $e \in K$ and $x \in K \implies x^{-1} \in K$.
Since $H$ is a group, the identity element $e$ is in $H$. Also, for any $x$ in $G$, $xex^{-1}=e\in H$ so for any $x\in G, x\in K$. In other words $G\subseteq K$, but also by definition, $K\subseteq G$. So $K=G$.