Assume I have a subset $X$ (not necessarily a subspace) of $\mathbb{F}_2^n$, of size $\leq 2^{n-1}$. It seems likely to me that there should exist a bijective linear transformation taking $X$ to a subset $Y$ contained in the $(n -1)$-dimensional subspace $V \subset \mathbb{F}_2^n$, being isomorphic to $\mathbb{F}_2^{n-1}$ (i.e. $V$ is the embedding of $\mathbb{F}_2^{n-1}$ inside $\mathbb{F}_2^n$). However, I'm finding this hard to argue, since almost all statements I've found that could be somewhat relevant for this is based on linear transformations between sub-spaces or affine spaces.
As per Marc van Leeuwen's comment below, an alternative statement of the problem, which is probably clearer, is the following. Assume $X$ spans the entire space $\mathbb{F}_2^n$ (since if not, the solution is trivial). What I'm looking for then, is a linear map $f$ whose restriction to $X$ is injective and equal to $Y$.
Any idea how one might attack this? I've been thinking about some sort of counting argument and the pigeonhole-principle, but I'm finding it tricky. A non-constructive proof would be just fine.
What you want to prove appears not to be true.
Note that if $X$ spans the whole space but one wants (as you do) to obtain that $Y=f(X)$ does not span the whole space, then $f$ cannot itself be injective (hence for dimension reasons bijective) since linear maps preserve spans. dimension reasons $f$ must have nonzero kernel. By just requiring the restriction of $f$ to $X$ to be injective, you want all elements of $X$ to be non-congruent modulo the kernel of $f$.
This means that the difference (equivalently sum) of any two distinct elements of$~X$ is forbidden to be in $\ker f$. But by an appropriate choice of $X$ one can arrange that this excludes every nonzero vector from $\ker f$, forcing $f$ to be injective which we saw it cannot be. For $n=3$, four vectors only give $\binom42=6$ differences which cannot exhaust $7$ nonzero vectors, however with $n=4$ one has ample space for a counterexample; even just $6$ well chosen vectors will do. Concretely take $V=\Bbb F_2^4$ and $$X=\{(0,0,0,0),~(0,0,0,1),~(0,0,1,0),~(0,1,0,1),~(1,0,0,0),~(1,1,1,0)\}. $$ One can easily check that all $15$ nonzero elements of the vector space can be written as a sum of two of these vectors. This means that any nonzero subspace contains the difference of two vectors, and so for a linear map $f:V\to V$ to restrict injectively to $X$, its kernel can contain only the zero vector. But then $f$ is an isomorphism and $Y=f(X)$ spans all of $V$; it is not contained in any subspace of dimension$~3$.