Mapping Normalizer to Automorphism I am squeezing my brain trying to understand this problem:

Let H be a subgroup of $G$, and denoted by $\operatorname{Aut}(H)$ the group of all automorphisms of $H. $ Define
$$\alpha : N_G(H) \to \operatorname{Aut}(H),\quad g \mapsto \alpha_g.$$
where $\alpha_g : H \to H, h \mapsto h^g.$ Show that $\alpha$ is a homomorphism and determine $\ker (\alpha).$

I understand that $N_G(H) = \{g \in G : Hg = gH \}$ and that $\operatorname{Aut}(H)$ is a set whose elements are automorphisms $\alpha_g : H \to H,$ But what I could not imagine is the mapping from $g$ (element of group $G$) to $\alpha_g$ (which in the most mundane term is actually a function.)
Additionally, I would love to see the synopsis of the solution. Any patient and friendly explanation befitting a person still in learning curve is very much appreciated. Thank you for your time and help. Happy holidays.
 A: $N_G(H)$ is a group — specifically, a subgroup of $G$. $\operatorname{Aut}(H)$ is also a group: it’s elements are functions — specifically, automorphisms of $H$ — and its operation is composition of functions. Thus, the function $\alpha$ that assigns to each $g\in N_G(H)$ the automorphism $\alpha_g\in\operatorname{Aut}(H)$ defined by $\alpha_g(h)=h^g$ for each $h\in H$ is simply a function from one group to another; don’t let the fact that the elements of $\operatorname{Aut}(H)$ are themselves functions get in your way.
Since $\alpha$ is a function from one group to another, it could conceivably be a homomorphism, and our task is to show that in fact it is one. In other words, we need to show that for any $g_0,g_1\in G$, $$\alpha_{g_0g_1^{-1}}=\alpha_{g_0}\circ\alpha_{g_1}^{-1}\;.\tag{1}$$
The two sides of $(1)$ are functions from $H$ to $H$; to show that they’re equal, we must show that 
$$h\alpha_{g_0g_1^{-1}}=h\left(\alpha_{g_0}\circ\alpha_{g_1}^{-1}\right)$$
for each $h\in H$. (Here I’m writing the function after its argument.) Let $h\in H$; then
$$\begin{align*}
h\alpha_{g_0g_1^{-1}}&=h^{g_0g_1^{-1}}\\
&=(g_0g_1^{-1})^{-1}h(g_0g_1^{-1})\\
&=g_1g_0^{-1}hg_0g_1^{-1}\\
&=g_1h^{g_0}g_1^{-1}\\
&=(h^{g_0})^{g_1^{-1}}\\
&=(h\alpha_{g_0})\alpha_{g_1^{-1}}\\
&=h(\alpha_{g_0}\circ\alpha_{g_1^{-1}})\;,
\end{align*}$$
and $\alpha$ is indeed a homomorphism.
The identity element of $\operatorname{Aut}(H)$ is the identity map $\operatorname{id}_H$ from $H$ to $H$, so 
$$\begin{align*}
\ker\alpha&=\{g\in N_G(H):\alpha_g=\operatorname{id}_H\}\\
&=\{g\in N_G(H):h^g=h\text{ for all }g\in H\}\;.
\end{align*}$$
With just a little more work you should be able to translate that definition into a familiar subgroup of $H$.
A: just assume that $H=G$ for now.  $G$ acts on itself by conjugation, and conjugation by an element is an automorphism of $G$.  if $\alpha_g:G\to G$ is conjugation by $g$, we have
$$
a_{g_1g_2}(h)=(g_1g_2)h(g_1g_2)^{-1}=\alpha_{g_1}(\alpha_{g_2}(h)),
$$
i.e. $\alpha:G\to\text{Aut}(G)$, $g\mapsto\alpha_g$, is a homomorphism (the group operation on automorphisms is composition of functions).  the kernel of $\alpha$ is the set
$$
\{g\in G : \alpha_g\equiv 1_G\},
$$
which is a way of describing $Z(G)$, the center of $G$ (the elements of $G$ that commute with everything).
Now you should be able to generalize this to a subgroup and it's normalizer.
A: You have the right idea:  $\alpha$ is indeed a function.  What that means is that for each element $g \in N_G(H) \le G$, we have a well-defined rule for associating an automorphism of the subgroup $H$, namely conjugation by $g$.
It's important that we only take elements $g$ from the normalizer of $H$, for those are the elements that fix $H$ setwise under conjugation.  (This is weaker than fixing $H$ pointwise; i.e. the centralizer of $H$.  More on this later).  In symbols,
$$
N_G(H) = \bigl\{ g \in G \mid gHg^{-1} = H \bigr\},
$$
so the function $\alpha_g$ maps $h \mapsto ghg^{-1} \in H$.
Now to check that $\alpha$ is a homomorphism, we have to show that $\alpha(g_1g_2) = \alpha(g_1)\alpha(g_2)$.  Note that the operation on the elements $g_1$ and $g_2$ is in the group $G$; whereas, the operation on the automorphisms $\alpha(g_1)$ and $\alpha(g_2)$ is composition of functions.
(Hover over the box to reveal the calculation.)

 $$ \alpha(g_1g_2)(h) = (g_1g_2) h (g_1g_2)^{-1} = g_1g_2 h g_2^{-1}g_1^{-1} = \alpha(g_1)(g_2 h g_2^{-1}) = \alpha(g_1)\bigl(\alpha(g_2)(h)\bigr) $$

What's the kernel?  An element $g \in \ker \alpha$ maps to the trivial automorphism on $H$ (the identity function).  In other words, $\alpha_g = \alpha_e$.  How do we characterize such $g$?

 $$ \alpha_g(h) = \alpha_e(h) \quad\Longrightarrow\quad ghg^{-1} = ehe^{-1} = h \text{ for all } h \in H \quad\Longrightarrow\quad g \in C_G(H)$$

