Extending a continuous map over projective space Let $X = P^{n-1}(\Bbb C)$ ($(n-1)$-dimensional projective space) with $n \geq 3$, and let $K \subset X$ denote a compact subset. I have a bijective, continuous map $\phi:K \to K$ which satisfies the condition
$$
\hat u + \hat v \supset \hat w \implies \phi(\hat u) + \phi(\hat v) \supset \phi(\hat w) \tag{1}
$$
for all $\hat u,\hat v,\hat w \in P^{n-1}(\Bbb C)$. Here, I am interpreting $\hat u \in P^{n-1}(\Bbb C)$ to be a $1$-dimensional subspace of $\Bbb C^n$, so that $\hat u + \hat v$ is the smallest subspace containing both $\hat u$ and $\hat v$.
The question is as follows: can I guarantee that $\phi$ will have an extension $\Phi:X \to X$ which satisfies (1) for all $\hat u,\hat v, \hat w \in X$? 
The motivation here is that, by the fundamental theorem of projective geometry, such a map $\Phi$ is necessarily induced by a semilinear map; it would be nice if I could guarantee that $\phi$ is induced by such a map. If anyone could point me in the direction of a theorem or reference that might be useful here, I'd appreciate it. 

Note: I've done my best to make this question self-contained and concise, and have therefore left out background that I felt was extraneous.  Feel free to comment if you feel that I should provide further context or clarification
 A: Take $n=2$. If $x$, $y$ are two distinct points of $\mathbb{P}^1$, the condition $x+y\supset z$ is satisfied for any $z$, so the condition on $\phi$ holds whenever   $\phi$ is injective, while $\phi$ has (in general) no semilinear extension if $K$ is big enough. 
A: For larger $n$ we can still make a counterexample, can't we? 
Let $S$ be a set of $n+1$ points in general position, and choose them to have rational coordinates. Any linear map which fixes these points must be the identity (on the projective space), and any semilinear map which fixes them must therefore be induced by a field automorphism, and must therefore fix every rational point. Now let $K$ consists of $S$ and two more rational points, and let $\phi:K\to K$ be the transposition interchanging these two points.
A: Your condition (1) means: if $\hat{u}$, $\hat{v}$, and $\hat{w}$ are linearly dependent, then so are $\widehat{\varphi(u)}$, $\widehat{\varphi(v)}$, and $\widehat{\varphi(w)}$. So $\varphi$ preserves linear dependence of $3$ vectors. But it doesn't detect linear dependence among $4$ or more vectors.
The other answers have given counterexamples where $K$ is a finite set of points. Basically, it's not hard to construct finite sets of points where no $3$ are linearly dependent, but there is a dependence among some $4$ or more of them.
I'll give a counterexample where $K$ is connected. Let $K$ be a rational normal curve, that is, the curve parametrized by $[s^{n-1} : s^{n-2} t : s^{n-3} t^2 : \dotsb : t^{n-1} ]$ in homogeneous coordinates on $X = \mathbb{P}^{n-1}$. Now $K$ doesn't have any trisecant lines, i.e., there are no $3$ points $\hat{u},\hat{v},\hat{w} \in K$ that are collinear in projective space (equivalently, $u,v,w \in \mathbb{C}^n$ are linearly dependent). So condition (1) is vacuously true. Then take $\phi$ to be any old map on $K$; then $\varphi$ satisfies (1). But if $\phi$ isn't semilinear then it doesn't extend to $X$.
