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.
From the comments above the proof is wrong because if $\langle v_1 \rangle \ldots \langle v_n\rangle$ are distinct it does not follow that all of these spaces are linearly independent.
Moreover the claim is false. We have that $GL(n,K)$ acts 3-transitively iff $n=2$ and hence $GL(3,K)$ does not act $3$ transitively.
To show that $GL(n,K)$ does not act $3$-transitively for $n\geq 3$ suppose that it was.
Then we can take $\langle v_1\rangle, \langle v_2\rangle , \langle v_3 \rangle \in K^n$ all distinct and linearly independent (as $n\geq 3$)
We can then also choose take $\langle v_1\rangle, \langle v_2\rangle, \langle v_1+v_2\rangle$ which are all distinct but no longer independent.
Then we have to have $g\in GL(n,K)$ with $v_1^g=\lambda_1 v_1$, $v_2^g=\lambda_2$ and $v_3^g=\lambda_3(v_1+v_2)$ but as $g$ is invertible this will give a contradiction