For what $n \in \mathbb{N}$ is $ \operatorname{Hol}(C_2^n)$ complete? Here $ \operatorname{Hol}$ stands for holomorph, and $C_2^n$ stands for direct product of $n$ isomorphic copies of $C^2$.
It for $n = 1$ it is not true, however if $n = 2$, then $ \operatorname{Hol}(C_2^n) \cong S_4$ is complete (proof that $ \operatorname{Hol}(C_2^n) \cong S_4$ can be found here: The holomorph of $Z_2 \times Z_2$). However, that is based on a specific property of $n = 2$ and does not help us in general case.
One can also easily see, that $ \operatorname{Hol}(C_2^n) = C_2^n \rtimes \operatorname{Aut}(C_2^n)$ is centerless for all $n > 1$, as it $ \operatorname{Aut}(C_2^n)$ acts both transitively and effectively on $C_2^n$.
However, determining for what $n$ all automorphisms of $ \operatorname{Hol}(C_2^n)$ are inner is a much harder task and I do not know how to solve it.
Any help will be appreciated.
Let $G={\rm Hol}(C_2^n) = C_2^n \rtimes {\rm Aut}(C_2^n) = C_2^n \rtimes {\rm GL}(n,2)$. You have discussed $n=1,2$, so let's assume that $n \ge 3$.
Let $\alpha$ be an outer automorphism of $G$. The subgroup $C_2^n$ is clearly characteristic in $G$, and hence fixed by $\alpha$.
The outer automorphism group of the simple group ${\rm GL}(n,2)$ has order $2$, and $A \mapsto (A^T)^{-1}$ defines an outer automorphism, for matrices $A$, where $A^T$ is the transpose of $A$. In the action on the natural module $C_2^n$, outer automorphisms interchange stabilizers of subspaces of dimension $r$ with those of dimension $n-r$. So they do not induce actions on $C_2^n$, and hence they do not extend to automorphisms of $G$.
So we may assume that $\alpha$ induces the identity on ${\rm GL}(n,2)$. Since its action on the natural module is absolutely irreducible, using Schur's Lemma we see that $\alpha$ must also induce the identity on $C_2^n$.
So $\alpha$ interchanges the complements of $C_2^n$ in $G$. In fact there is a bijection between these complements and the group of automorphisms of $G$ that induce the identity on ${\rm GL}(n,2)$, where inner automorphisms correspond to those complements that are conjugate to the principal complement in $G$. So the group of outer automorphisms of $G$ is isomorphic to the cohomology group $H^1({\rm GL}(n,2),C_2^n)$.
It is known that this cohomology group has order $2$ for $n=3$ and is trivial for $n>3$. So $G$ is complete if and only if $n \ge 4$ (and if $n=2$).