If we let $K$ be a field and $GL(n,K)$ act by right multiplication on the $1$-dim subspaces of $K^n$.
Then if we take $\langle v_1 \rangle, \ldots \langle v_n \rangle \in K^n$ distinct and $\langle w_1 \rangle, \ldots \langle w_n \rangle \in K^n$ again distinct.
Then $v_1, \ldots , v_n$ forms a basis of $K^n$ as does $w_1, \ldots , w_n$.
So we know that there is an invertible map $f:K^n \rightarrow K^n$ for which $v_i^m = w_i$ for $1 \leq i \leq n$ with $m \in GL(n,K)$ and moreover $\langle v_n \rangle ^m = \langle w_i \rangle$ hence $GL(n,K)$ acts n-transitively.
Is the above correct?
Thanks for any help.