Let $H,K$ be group where $\phi: K \rightarrow \operatorname{Aut}(H)$ is a homomorphism. Also, let $G=H \rtimes_{\phi} K$.
Show that $C_H(K) = N_H(K)$
Proof: Let $h\in N_H(K)=\{h\in H: hKh^{-1}=K\}$. Then for any $k\in K$, $\exists \; k' \in K$ where $hkh^{-1}=k'$. Let $\pi_{k'}(h)$ be an action by left conjugation that affords $\phi$. Then $\pi_{k'}(h)=k'hk'^{-1}=(hkh^{-1})h(hkh^{-1})^{-1}=hkh^{-1}hhk^{-1}h^{-1}=(hk)h(hk)^{-1}$. Thus $hk=k'$ and since $H\cap K=1$, then $h=1$. Thus $N_H(K)=1$ and $C_H(K)=1$ (as it is a subgroup of $N$).
Is this proof correct? Huge Thanks goes out to anyone who helps me out here!
With the notation in $\phi_k = \phi(k) \in \operatorname{Aut} H$, the multiplication is given by $$ (h_1, k_1) (h_2, k_2) = (h_1\, \phi_{k_1}(h_2), k_1 k_2). $$ You have probably already done the calculations to show that $H$ and $K$ embed in $G = H \rtimes_\phi K$ as $$ \{ (h, 1) \mid h \in H \} \quad \text{and} \quad \{ (1, k) \mid k \in K \}. $$ Also, you've likely worked out that $$ (h, k)^{-1} = ( \phi_{k^{-1}}(h^{-1}), k^{-1}). $$ Now choose $h \in H$ and $k \in K$ and conjugate in $G$: $$ (h, 1) (1, k) (h, 1)^{-1} = (h, k) (h^{-1}, 1) = (h \phi_k(h^{-1}), k). $$ If $h \in N_H(K)$, then $(h \phi_k(h^{-1}), k) = (1, k')$ for some $k' \in K$. This implies that $k' = k$, so $h \in C_H(K)$.