Consider a basis $\left\{ {{w_1},{w_2},...,{w_s}} \right\}$ to GF($2^s$) (for some $s \ge 2$) over GF($2$). I am trying to show the existence of an element $g \in$ GF($2^s$) such that each of the $s-1$ sets $$\left\{ {g{w_1},{w_1},{w_2},...,{w_{s - 1}}} \right\},\left\{ {g{w_1},g{w_2},{w_1},{w_2},...,{w_{s - 2}}} \right\},...,\left\{ {g{w_1},g{w_2},...g{w_{s - 1}},{w_1}} \right\}$$ is a basis to GF($2^s$) over GF($2$).
The observation I made so far is that by discarding $g$ values that violate the independence of the sets above, we actually discard additive subgroups of the field. To see this, denote by $\left\langle {{\alpha _1},{\alpha _2},...,{\alpha _s}} \right\rangle $ the span (set of linear combinations) of the field elements ${{\alpha _1},{\alpha _2},...,{\alpha _s}}$ with coefficients taken from GF($2$): $$\left\langle {{\alpha _1},{\alpha _2},...,{\alpha _s}} \right\rangle = \left\{ {\sum\limits_{i = 1}^s {{\beta _i} \cdot {\alpha _i}:{\beta _i} \in \left\{ {0,1} \right\}} } \right\}.$$
To ensure linear independence for $\{ {g{w_1},{w_1},{w_2},...,{w_{s - 1}}}\}$, $g$ must not belong to the subgroup $${\mathcal{S}_1} = \left\langle {{w_1},{w_2},...,{w_{s - 1}}} \right\rangle /{w_1} = \left\{ {\sum\limits_{i = 1}^{s - 1} {{\beta _i} \cdot {w_i}} :{\beta _i} \in \left\{ {0,1} \right\}} \right\}/{w_1}.$$ (division is element-wise).
That is, $g$ belongs to a coset of $\mathcal{S}_1$. In a similar fashion, to ensure linear independence of the set $\{ {g{w_1},g{w_2},{w_1},{w_2},...,{w_{s - 2}}}\}$, $g$ must belong as well to cosets of the subgroups $\mathcal{S}_2 = {\left\langle {{w_1},{w_2},...,{w_{s - 2}}} \right\rangle /{w_2}}$ and $\mathcal{S}_3 = \left\langle {{w_1},{w_2},...,{w_{s - 2}}} \right\rangle /\left( {{w_1} + {w_2}} \right)$ and so on. In general, set $i$ ($i=1,2,...,s-1$) induces the constraint that $g$ belongs to $2^{i-1}$ cosets of $2^{s-i}$ elements each.
I am trying to show that the intersection of these cosets (note that a coset leader needs to be specified for each coset) is non-empty.
Empirically, I found that using a polynomial basis (regardless of the primitive polynomial in use) and up to $s=12$, there are exactly $2$ such $g$ values for each $s$, where one is the multiplicative inverse of the other. That is, the intersection of the cosets above is of cardinality $2$.
Example ($s=3$)
For $s=3$ we need to find $g$ such that the elements in both sets $\left\{ {g{w_1},{w_1},{w_2}} \right\}$ and $\left\{ {g{w_1},g{w_2},{w_1}} \right\}$ are linearly independent. That is, $g$ must belong to the intersection of certain cosets of the subgroups $\left\langle {{w_1},{w_2}} \right\rangle /{w_1} = \left\{ {0,{w_1}/{w_1},\left( {{w_1} + {w_2}} \right)/{w_1},{w_2}/{w_1}} \right\}$, $\left\langle {{w_1}} \right\rangle /{w_2} = \left\{ {0,{w_1}/{w_2}} \right\}$ and $\left\langle {{w_1}} \right\rangle /\left( {{w_1} + {w_2}} \right) = \left\{ {0,{w_1}/\left( {{w_1} + {w_2}} \right)} \right\}$.