A version of Robin Chapman's answer that might be easier to verify:
The additive group $H$ of a field of order $2^{n}$ has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism. The semi-direct product $G$ is called $AΓL(1,2^{n})$. It has order $(2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n}$, and for $n ≥ 2$ it is its own automorphism group.
It has the additive group $H$ of the field as a normal subgroup. However, as just an additive group, $H$ has automorphism group all n×n matrices over the field with 2 elements, which has size $(2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) ≥ 2^{(n−1)^{2}}$.
For large $n$, $|Aut(G)| = (2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n} ≤ 2^{(n−1)^{2}} ≤ (2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) = |Aut(H)|$
In particular:
For $n = 4$, $G = AΓL(1,16)$, $Aut(G) = G$ has order $960$, and $Aut(H) = GL(4,2)$ has order $20160$.
For $n = 10$, $G = AΓL(1,1024)$, $Aut(G) = G$ has order $10475520$, and $Aut(H) = GL(10,2)$ has order $366440137299948128422802227200$.
In other words, $Aut(H)$ can be enormously bigger than $Aut(G)$.
This is reasonably important in finite group theory:
$G$ is a very rigid group with lots of structure. Because it contains all the "automorphisms" of the field, the group itself determines the field. An automorphism of the group will have to be an automorphism of the field, and we've already listed them all. Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.
$H$ is a very floppy group to which you can do nearly anything. Without the maps encoding scalar multiplication, $H$ no longer remembers the field that defined it. It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism. $H$ has lost its structure.
I think $H$ is the canonical example of a horribly structureless group. Groups like $GL$ and $AΓL$ are pretty standard examples of groups with very clear structure. The symmetric group is very similar. Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.
Even without the classification of finite simple groups, quite reasonable bounds are known, for example in work of P.M. Neumann. If the group $G$ can be generated by $r$ but no fewer elements, then no automorphism of $G$ can fix the $r$ given generators, so there are at most $\prod_{j=1}^{r} (|G|-j)$ different automorphisms of $G,$ since the $r$ generators must have distinct images, none of which is the identity. As P.M. Neumann has observed, $G$ can always by generated by ${\rm log}_{2}(|G|)$ or fewer elements, so we have $r \leq \lfloor {\rm log}_{2}(|G|) \rfloor .$ For $|G| >4,$ this always gives a strictly better bound for the size of ${\rm Aut}(G)$ than $(|G|-1)!.$ For large $|G|,$ it is much better. Using the classification of finite simple groups, much better bounds are known.
Later edit: Perhaps I could outline Neumann's argument, since it is quite elementary, and I do not remember a reference: Let $\{x_1, x_2, \ldots, x_r \}$ be a minimal generating set for $G$ and let $G_i = \langle x_1, x_2, \ldots, x_i \rangle $ for $i >0,$ $G_{0} = \{ e \}.$ Then for $1 \leq i \leq r,$ we have $|G_i| > |G_{i-1}|$ by minimality of the generating set. Furthermore, $|G_i|$ is divisible by $|G_{i-1}|$ by Lagrange's theorem, so $|G_i| \geq 2|G_{i-1}|.$ Hence $|G| = |G_r| \geq 2^r.$
Best Answer
This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $\operatorname{Aut}(H)$, thus $\left\lvert N_G(H)/C_G(H)\right\rvert$ divides $\left\lvert \operatorname{Aut}(H)\right \rvert$. But $\left\lvert N_G(H)/C_G(H)\right\rvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $\gcd(|G|,\left\lvert \operatorname{Aut}(H)\right \rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) \leq N_G(H)$ we get $N_G(H)=C_G(H)$.