Sets from $(F_2)^n$ which are not fixed by any non-identity isomorphism This is a followup question to the discussion in the comments of
Sets which are not fixed by any non-identity isomorphism
So consider a finite $n$-dimensional vector space $V$ over $F_2$. For which such $n$ does there exist a subset $W$ of $V$ such that $T(W)\neq W$ for any non-identity isomorphism $T:V\to V$.
$n=1$ is trivial, for $n=2,3$ one can see no subset works. For $n=4,5$ perhaps the set $\{e_1,e_2,e_3,e_4,e_1+e_2,e_1+e_2+e_3,e_1+e_2+e_3+e_4,e_1+e_4\}$, with the $e_i$s being a basis, is worth checking. (Or possibly I'm missing something trivial).
It seems interesting to me how one goes about proving if such sets exist or do not exist for all large $n$.
 A: Now that Peter Mueller has shown that solutions exist for large values of $n$, 
I believe it is worthwhile to give a more explicit construction of a solution. 
The idea is to work by induction on $n$, seeing $(F_2)^n$ as a linear subspace of 
$(F_2)^{n+1}$ through the canonical injection $(x_1,\dots,x_n) \mapsto (x_1,\dots,x_n,0)$.
Now, let us assume that, for some integer $n>2$, there exists a subset $W_n$
of $(F_2)^n$ for which the identity is the sole automorphism of $(F_2)^n$ that leaves it invariant. Then, we construct a subset $W_{n+1}$ of $(F_2)^{n+1}$ as follows:
If $|W_n|>2^{n-1}$, we define $W_{n+1}$ as the union of $W_n$ with the 
set consisting of the sole vector $e_{n+1}:=(0,\dots,0,1)$; 
Otherwise, we define $W_{n+1}$ as the union of $W_n$ with the 
set consisting of all the vectors $(x_1,\dots,x_n,1)$ with $(x_1,\dots,x_n) \neq 0$. 
I contend that $W_{n+1}$ is a solution in any case. From there, 
if one has an explicit solution for an integer $p>3$, then one obtains an explicit
solution for all larger values of $p$ by noting that, 
in the above construction $|W_{n+1}|<2^n$ whenever
$|W_n|>2^{n-1}$, whereas $|W_{n+1}|>2^n$ whenever
$|W_n|\leq 2^{n-1}$ (this comes from the observation that $1<|W_n|<2^n-1$). 
Now, let us prove the claimed statement. Actually, only the first case needs to be considered since the complementary subset of a solution in the whole vector space is another solution. So, let us assume that $|W_n|>2^{n-1}$ and let us consider an
automorphism $f$ of $(F_2)^{n+1}$ that leaves $W_{n+1}$ invariant. 
We set $\mathcal{H}:=e_{n+1}+(F_2)^n$. 
Here is the key point:

$\mathcal{H}$ is the sole non-linear hyperplane of $(F_2)^{n+1}$ that has at most one common point with $W_{n+1}$. 

Proof :
Indeed, let $\mathcal{G}$ be a non-linear hyperplane of $(F_2)^{n+1}$ that differs from 
$\mathcal{H}$; then $\mathcal{F}:=\mathcal{G} \cap (F_2)^n$ is a non-linear hyperplane of 
$(F_2)^n$, and we claim that it must have several common points with $W_n$:
if indeed this were not the case, as $|W_n|>2^{n-1}$, the set $W_n$
should be the union of the translation vector space $F$ of $\mathcal{F}$ with $\{y\}$
for some $y \in \mathcal{F}$; but then at least one non-identity automorphism of $(F_2)^n$
leaves $W_n$ invariant: one simply takes an arbitrary non-identity automorphism of $F$ and extends it to $(F_2)^n$ by requiring that it should fix $y$.

From there the proof is easy: using the above result, one sees that $f$ must leave
$\mathcal{H}$ invariant, whence $f$ fixes $W_n \cap \mathcal{H}=\{e_{n+1}\}$ and it leaves
$(F_2)^n$ invariant as it is the translation vector space of $\mathcal{H}$. 
By induction, the second result yields that the restriction of $f$ to $(F_2)^n$ is the identity. Therefore, $f$ is the identity. 
A: For $n=4$ it cannot work: Suppose that no nontrivial element of $G=GL(n,2)$ fixes an element of $W$. Then $V$ has at least $\lvert G\rvert$ subsets of size $k=\lvert W\rvert$, because all the sets $g(W)$, $g\in G$, are distinct.
So a necessary condition is $\lvert GL(n,2)\rvert\le\binom{2^n}{k}$. However, $\lvert GL(4,2)\rvert=20160>12870=\binom{16}{8}\ge\binom{16}{k}$.
For $n\ge5$ this obstruction does not hold anymore. Indeed, experiments show that for $n=5$ there are examples with $k=16$, and for $n\ge6$, apparently most random subsets $W$ of size $2^{n-1}$ work.
An easy counting argument shows that for $n$ sufficiently big, indeed such a set $W$ exists: Let $f$ be the number of fixed point of a nonzero element from $GL(n,2)$. Then $g$ has at most $(2^n-f)/2$ cycles of length $\ge2$ on $V$. So all together, $g$ has at most $2^{n-1}+f/2$ cycles. But $f\le 2^{n-1}$, so any nontrivial $g$ has at most $2^{n-1}+2^{n-2}=2^n-2^{n-2}$ cycles. Each invariant subset of $V$ is a union of such cycles. So there are at most $2^{2^n-2^{n-2}}$ invariant subsets.
Set $M=\{(w,g)|g(w)=w\}$, where $w$ runs through the subsets of $V$, and $g$ through $G\setminus\{1\}$. By what we saw, $\lvert M\rvert\le (\lvert G\rvert-1) 2^{2^n-2^{n-2}}$. On the other hand, if there is no set $W$, then each $w$ is fixed by at least one $g\ne1$, hence $\lvert M\rvert\ge 2^{2^n}$.
We obtain $\lvert G\rvert\gt 2^{2^{n-2}}$. On the other hand, clearly $\lvert G\rvert\le 2^{n^2}$, a contradiction for $n\ge8$. So the cases $n=5,6,7$  probably have to be done by hand (or computer). Indeed, in these cases $W$ exists.  
A: 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. 
