To complete the preceding answers, here's an explicit solution for $n=5$:
One denotes by $B$ the set of all vectors of the standard basis of $(F_2)^5$,
by $\mathcal{H}$ the hyperplane of $(F_2)^5$ defined by the equation
$\sum_{k=1}^5 x_k=1$, and one sets $K=\{(0,1,1,0,0),(0,0,1,1,0),(1,1,1,1,0),(1,1,1,0,1)\}$.
Then, $W:=K \cup (\mathcal{H} \setminus B)$ is a solution.
What follows is a rough sketch of proof with no resort to computer verifications.
First of all, one checks that, if we denote by $\mathcal{H}'$ the hyperplane defined by the equation $x_3=1$, then $\mathcal{H}$ and $\mathcal{H}'$ are the only non-linear hyperplanes that contain exactly $11$ elements of $W$. Thus, every automorphism of $(F_2)^5$ that leaves $W$ invariant must either leave $\mathcal{H}$ and $\mathcal{H}'$
invariant or swap them.
The next step consists in proving that the sole automorphism that leaves $W$ and $\mathcal{H}$ invariant is the identity: this is easily done by noting that such an automorphism $f$ must leave $B$ invariant, whence it has the form
$(x_1,\dots,x_5) \mapsto (x_{\sigma(1)},\dots,x_{\sigma(5)})$ for some permutation $\sigma$
of $\{1,\dots,5\}$; then, using the fact that $f$ must also leave $K=W\setminus \mathcal{H}$ invariant, one easily shows that $\sigma=id$.
It remains to prove that there is no automorphism $f$ that swaps $\mathcal{H}$ and $\mathcal{H}'$ and leaves $W$ invariant: assume on the contrary that such a map $f$
exists. Then, $f$ must swap $B=\mathcal{H} \setminus W$ and $B':=\mathcal{H}' \setminus W=\bigl\{(0,0,1,0,0),(1,0,1,0,0),(0,0,1,0,1),(0,1,1,1,1),(1,0,1,1,1)\bigr\}$.
Now, $(1,1,1,1,1)$ is the sum of all the vectors in $B$, and also the sum of all the
vectors in $B'$, whence $f$ fixes it.
Moreover, $f$ fixes $e_3:=(0,0,1,0,0)$ as it is the sole vector of $B \cap B'$.
On the other hand, as $f^2$ leaves $W$ and $\mathcal{H}$ invariant, we know from the first step above that $f^2=id$. Finally, a contradiction is obtained by
looking at the bijection from $B \setminus \{e_3\}$ to $B' \setminus \{e_3\}$
induced by $f$ and, knowing that $f(e_3)=e_3$, by checking the equality $f^2=id$.
One can test all the $24$ cases (tedious).
Alternatively, one can reason as follows:
one computes the table of all the sums $b+b'$ with $b \in B \setminus \{e_3\}$ and $b' \in B' \setminus \{e_3\}$, and then one uses the following principles to
dicard all the possible bijections from $B \setminus \{e_3\}$ to $B' \setminus \{e_3\}$.
- $\mathop{im}(f-id) \subset \ker(f-id)$, and therefore $\mathop{rk}(f-id)\leq 2$ by the rank theorem.
- The union $\{b+f(b) \mid b \in B \setminus \{e_3\}\} \cup \{(1,1,1,1,1),(0,0,1,0,0)\}$
is included in $\ker(f-id)$, whereas $\{b+f(b) \mid b \in B \setminus \{e_3\}\}$
is included in $\mathop{im}(f-id)$.
One intermediate result is to prove that the map $b \mapsto b+f(b)$ takes only two values on $B \setminus \{e_3\}$, each attained twice. Then, the contradiction is easily found by table-chasing.