Followup to question concerning $N_G(H) / C_G(H) \cong B \leq \mathrm{Aut}(H)$ 
Prove $H \leq G$ implies $N_G(H) / C_G(H) \cong B \leq \mathrm{Aut}(H)$ where $B$ is some subgroup of $\mathrm{Aut}(H).$

I did see this question:
For $H \leq G$, showing that $N_G(H)/C_G(H) \leq \text{Aut}(H)$
which has been helpful. My question now is with the Proposition 13 that the question makes reference to:

$\textbf{Proposition 13: }$ Let $H$ be a normal subgroup of the group $G$. Then $G$ acts by conjugation on $H$ as automorphisms of $H$. More specifically, the action of $G$ on $H$ by conjugation is defined for each $g \in G$ by $$h\mapsto ghg^{-1} \text{ for each $h \in H$.}$$ For each $g \in G$, conjugation by $g$ is an automorphism of $H$. The permutation representation afforded by this action is a homomorphism of $G$ into $\mathrm{Aut}(H)$ with kernel $C_G(H)$. In particular, $G/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$.

Specifically, my question is what do they mean by 

"The permutation representation afforded by this action..."?

I see that the conjugation map stated in the proposition above is an automorphism of $H$ and so it just permutes the elements of $H$ but I don't know what the permutation representation is or means.
My goal is, I think at least for now, to prove this proposition and the referred question above states that the result I need to show follows from there.
 A: $\textbf{Proof of Proposition 13}$: As $H$ is normal in $G$, $ghg^{-1}\in gHg^{-1}=H$ and so each $ghg^{-1}\in H$. If $h,k \in H$ then
$$f_g(hk)=g(hk)g^{-1} \text{ and } f_g(h)f_g(k)=(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$$
so $f_g(hk)=f_g(h)f_g(k)$ so $f$ is a homomorphism. If $h \in H=gHg^{-1}$ then $h=gh'g^{-1}$ for some $h' \in H$ and so
$$f_g(h')=gh'g^{-1}=h$$
so $f_g$ is a surjection. We have that
\begin{align*}
\mathrm{Ker}(f_g) &= \{ h \in H: f_g(h)=e_H \} \\
                &= \{ h \in H: ghg^{-1}=e_H \} \\
                &= \{ h \in H: gh=e_Hg \} \\
                &= \{ h \in H: gh=ge_H \} \\
                &= \{ h \in H: h=e_H \} \\
                &= \{ e_H \}
\end{align*}
so $f_g$ is injective. We conclude that $f_g$ is an isomorphism and as $f_g:H \rightarrow H$, we have $f_g$ is an automorphism. 
Let $f$ be as above. If $g,k \in G$ then
$$f(gk)=f_{gk} \text{ and } f(g) \circ f(k) = f_g \circ f_k.$$
For any $h \in H$, $f_{gk}(h)=(gk)h(gk)^{-1}$ and
$$(f_g \circ f_k)(h)=f_g( f_k(h) ) = f_g(khk^{-1})=g(khk^{-1})g^{-1}=(gk)h(gk)^{-1}=f_{gk}(h)$$
so $f_{gk} = f_g \circ f_k$ meaning
$$f(gk)=f(g) \circ f(k).$$
Therefore $f$ is a homomorphism. We have then
\begin{align*}
\mathrm{Ker}(f) &= \{ g \in G : f(g)=\mathrm{id}_H \} \\
                &= \{ g \in G: f_g = \mathrm{id}_H \}
\end{align*}
and so if $g \in \mathrm{Ker}(f)$ then for all $h \in H$, $f_g(h)=ghg^{-1}=h$ and so $gh=hg$ meaning that $g \in C_G(H)$. Further if $g \in C_G(H)$ then $f_g(h)=ghg^{-1}=hgg^{-1}=h=\mathrm{id}_H$ so $g \in \mathrm{Ker}(f)$. We have shown then
$$\mathrm{Ker}(f)=C_G(H).$$
Therefore, we can take the function $j: G \rightarrow \mathrm{Im}(f)$ by $j(g)=f(g)$. This function $j$ is a surjective homomorphism and so by the first isomorphism theorem we have that since $\mathrm{Ker}(f)=\mathrm{Ker}(j)$ that
$$G/C_G(H) \cong \mathrm{Im}(f)$$
and we know that $\mathrm{Im}(f) \leq \mathrm{Aut}(H)$ proving what was needed.
$\textbf{Proof of desired thing: }$ We have that $H$ is a normal subgroup of $N_G(H)$ and so from above  $N_G(H)/C_{N_G(H)} \cong B \leq \mathrm{Aut}(H)$ and further $C_{N_G(H)}(H)=C_G(H)$ proving the statement.
