That's what I thought too. This should work:
Interpret any binary vector $(x_{i})_{i=0}^{n-1}$ as $\sum_{i=1}^{n-1}x_{i}t^{i}$, where $t$ is a root of $f(x)=\frac{x^{n}+1}{x+1}$ in an algebraic closure of $\mathbb F_{2}$ (in other words, we consider the map $(x_{i})\mapsto\sum x_{i}t^{i}$, which is a group homomorphism from an $n$-dimensional vector space over $\mathbb F_{2}$ onto the additive group of the field $\mathbb F_{2}(t)$).
The $k$'th rotation of our vector $(x_{i})$ will then represent $\sum x_{i}t^{k+i}=t^{k}\cdot\left(\sum x_{i}t^{i}\right)$. If the rotations are not linearly independent, then a sum of several of them is equal to zero, and through the homomorphism described above this means $p(t)\cdot q(t)=0$ for polynomials $p,q\in\mathbb F_{2}[x]$ of degree $\le n-1$. One of these ($q$, for example) is just $\sum x_{i}x^{i}$, so $f$ does not divide it (because not all $x_{i}$ are $1$). Since $f|pq,\ f$ must divide $p$, and this means that $p=f$, i.e. the sum of all of our $n$ rotations of the initial vector must be zero. This sum, however, is equal to the vector having all entries equal to $1$, because the initial vector has an odd number of $1$'s. We have a contradiction.