Sets which are not fixed by any non-identity isomorphism Consider a finite dimensional vector space $V$ over a field (finite or infinite but big enough). I am looking for a subset $W$ of $V$ such that for any bijective but non-identity linear map $T: V \longrightarrow V$, $T(W) \not = W$. Is it always possible to find such a set ? 
 A: I will modify the construction of A.B. a bit; so it works for fixed field and arbitrary dimension of $V$.
Choose a basis $e_i$ in $V$ and consider the set $W$ 
which includes $e_i$, $2{\cdot} e_i$ for each $i$ and $e_i+2\cdot e_j$ for $i<j$.
Note that if $f(W)=W$ then the basis goes to basis
(these are the only elements $w\in W$ such that $2\cdot w\in W$).
Further note that the order of basis survives; i.e., $W$ is the set you want.
A: If the cardinality of the field is much larger than the dimension of $V$ then:
Start with a basis $B=\{v_1,...,v_n\}$ and then take the set $S=B\cup \{\alpha_1v_1,\alpha_2v_2,...,\alpha_nv_n\}$ (I guess in finite characteristic you need to choose the scalars carefully). 
If a linear transformation $T$ fixes $S$, then for some $v_i$, $T(v_i)\in\{\alpha_iv_i,v_j,\alpha_jv_j\}$.
Therefore, $T(\alpha_iv_i)\in\{\alpha_i^2v_i,\alpha_iv_j,\alpha_i\alpha_jv_j\}$. If $\alpha_i^2\neq 1$, $\alpha_i\neq \alpha_j$ and $\alpha_i\alpha_j\neq 1$ for all $i,j$ then this seems enough.
EDIT: (I'll modify Anton Petrunin's construction a bit to answer Anton Klyachko's question about $F_3$)
For the field of 3 elements, assuming $n>2$: Choose a basis $e_i$ and take $W$ which includes $e_i,2e_i$ for each $i$, $e_i+2e_j$ for $i<j$, and $\Sigma_{i=1}^{n} e_i$. Assume by contradiction that $T$ is a non-trivial automorphism of $V$ which fixes $W$.
If $T(e_i)=e_j+2e_k\in W$ then $j<k$ so $T(2e_i)=e_k+2e_j\notin W$, a contradiction. If $T(e_i)=\Sigma_{i=1}^{n} e_i$ then $T(2e_i)=2\Sigma_{i=1}^{n} e_i\notin W$. So for each $i$, $T(e_i)\in\{e_j,2e_j\}$.
Since if $e_i\neq e_j$ then $T(e_i)\neq \alpha T(e_j)$, we get that $T(\Sigma_{i=1}^{n} e_i)=\Sigma_{j=1}^{n}\alpha_j e_j\in W$ with $\alpha_j\in\{1,2\}$. Therefore, all the $\alpha_j$s must be 1.
That means that $T$ makes a permutation of the $e_i$s, and we get a contradiction by Anton's explanation.
A: Here is a method that works for all fields with more than $2$ elements. 
I will denote the underlying field by $F$ and assume that $|F|>2$. 
The idea is to use an induction to construct a solution
for $n$-dimensional spaces. I will consider $F^n$ as a linear subspace of 
$F^{n+1}$ by using the embedding $(x_1,\dots,x_n) \mapsto (x_1,\dots,x_n,0)$. 
Assume that we have a subspace $W_n$ of $F^n$ that is a solution, and 
consider the subspace $W_{n+1}$ of $F^{n+1}$ defined as the union of $W_n$
with the set of all vectors $(x_1,\dots,x_n,1)$ with $(x_1,\dots,x_n) \neq 0$. 
I contend that, unless $n=1$ and $|F|=3$, the subset $W_{n+1}$ is a solution for the 
space $F^{n+1}$. Indeed, let $f$ be an automorphism of $F^{n+1}$ that leaves $W_{n+1}$
invariant. Unless $n=1$ and $|F|=3$, one checks that the affine hyperplane $\mathcal{H}$ with equation $x_n=1$ is the sole non-linear affine hyperplane of $V$ such that 
$\mathcal{H} \setminus W$ contains at most one point. Thus, $\mathcal{H}$ is stable under $f$. It follows that $f$ fixes $(0,\dots,0,1)$, as it is the only point of $\mathcal{H} \setminus W$, and on the other hand the translation vector space $H$ of $\mathcal{H}$ - associated with the equation $x_n=0$ - is stable under $f$. By induction, $f$ is the identity on $H$. As $f$ is linear, one concludes that it is the identity. 
If $|F|>3$, this method yields a solution in every situation as one can start from 
$W_1:=\{1\}$. If $|F|=3$, there is no solution for $n=2$ as was noted by Harry Altman, 
but there is one for $n=3$: one sets $\mathcal{H}:=\{(x_1,x_2,1) \mid (x_1,x_2) \in F^2\}$ and
 $W_3=\{(1,0,0)\} \cup (\mathcal{H} \setminus \bigl\{(0,0,1),(1,0,1),(0,1,1)\}\bigr\}$. 
Then, one can check that $W_3$ is a solution for $F^3$: 
indeed, one sees that $\mathcal{H}$
is the sole affine hyperplane of $F^3$ such that $|\mathcal{H} \setminus W_3|=3$.
Let $f$ be an automorphism of $F^3$ that leaves $W_3$ invariant. Then, $f$
also leaves $\mathcal{H}$ invariant, and it must induce an affine automorphism of 
it that permutes the three points $(0,0,1)$, $(1,0,1)$ and $(0,1,1)$, 
and it must induce an automorphism of $F^2$ that fixes $(1,0,0)$ (as $F^2$
is the translation vector space of $\mathcal{H}$).
Note that $f$ must fix $(0,0,1)+(1,0,1)+(0,1,1)=(1,1,0)$, 
whence $f$ induces the identity on $F^2$. It follows that the restriction of $f$
to $\mathcal{H}$ is a translation, and from there it is easily checked that 
this translation is the identity by using the fact that $f$ leaves $\{(0,0,1),(1,0,1),(0,1,1)\}$ invariant. Thus, $f$ is the identity and $W_3$ is a solution, as claimed. 
With the above method, one obtains a solution for all $n\geq 3$ if $|F|=3$. 
