Let $R = \mathrm{GF}(q)$, $S = \mathrm{GF}(q^n), \ n\geq 2$ be extension of $R$, $h$ be a primitive element of $S$. I want to count or estimate the number $N$ of bases of the following form.
Let $$\vec{\beta} = (\beta\_0,\beta\_1,\ldots,\beta\_{n-1})$$ be a basis of the space $\_RS$ with the property that there exists $k\in\{1,2,\ldots,n-1\}$ such that
$$
\vec{\beta'} = (\beta\_0,\ldots,\beta\_{k-1},h\beta\_k,\ldots,h\beta\_{n-1}) 
$$
is a basis of the space $\_RS$. 


It is easily to see that $N\geq (q^n-q)\prod\_{j=1}^{n-1}(q^n-2q^{j}+q^{j-1})$. This estimation obtained from the case $k = n-1$ 


It is also easily to see, that the basis of the form $(e,h,\ldots, h^{n-1})$ has the desirable property.
My another question is: is there another approches for construction such bases?



Bases of the special form
Let me estimate the number $N\_k$ of such bases for a fixed $k$. *Aside remark.* The number of bases for $k=k\_0$ equals the number of those for $k=n-k\_0$, as the bijection $(\beta\_0,\dots,\beta\_{n-1})\to (\beta\_{n-1},\dots,\beta\_{k\_0},h^{-1}\beta\_{k\_0-1},\dots,h^{-1}\beta\_0)$$ shows.


So let us fix any $k$. The first $k$ elements can be chosen in $\prod\_{i=0}^{k-1}(q^n-q^i)$ many ways. Each next element $\beta\_i$ should lie outside $V\_i=\langle \beta\_0,\dots,\beta\_{i-1}\rangle$, as well as outside $V\_i'=\langle h^{-1}\beta\_0,\dots,h^{-1}\beta\_{k-1},\beta\_k,\dots,\beta\_{i-1}\rangle$. If $d\_i=\dim(V\_i\cap V\_i')$, then $\beta\_i$ can be chosen in $q^n-2q^i+q^{d\_i}$ many ways; notice that $i\geq d\_i\geq \max(i-k,2i-n)$. Thus
$$
 N\_k\geq \prod\_{i=0}^{k-1}(q^n-q^i)\prod\_{i=k}^{n-1}(q^n-2q^i+q^{\max(i-k,2i-n)}).
$$


This estimate can be improved by noticing that $d\_i\geq d\_{i-1}+1$. So, e.g., for $k=1$, if we choose $\beta\_1=(h^{-1}+1)\beta\_0$, then $V\_2=V\_2'$, so $d\_i=i$ for $i\geq 2$. But it seems appropriate to choose large $k$...


Notice that for $k=n-1$, the estimate gives
$$
 N\_{n-1}\geq (q^n-2q^{n-1}+q^{n-2}).\prod\_{i=0}^{n-2}(q^n-q^i),
$$
which is $1-\frac1q$ of all bases. Do you need much sharper bounds?


