Let $q$ be a power of a prime and $S \subseteq \mathrm P^2 \mathbf F^q$ such that
$$ \forall g \in \operatorname{PGL}(3,q), gS \cap S \neq \emptyset.$$
Can it be that $\vert S \vert < 1+q$ ?


(I asked a version of this question few years ago [here](https://math.stackexchange.com/questions/2073541/sets-smaller-than-a-line-that-cannot-be-disjointed-from-themself-by-projective-a) on math-stackexchange, but did not get an answer.)



Smallest subset in $P^2 \mathbf F_q$ which cannot be disjointed from itself by a homography
Assume that $|S|\leqslant q$. Choose a random projective map $g$. The probability that $gs=t$ for fixed $s,t\in S$ equals $1/|\mathrm P^2 \mathbf F^q\|=1/(q^2+q+1)$, so the sum over all pairs $(s,t)\in S\times S$ is strictly less than 1 and there exists $g$ such that $gs\ne t$ for all pairs.


