In other words, if $\Sigma$ is generic, there does not exist a non-identity automorphism $h$ for which $|\Sigma\cap h(\Sigma)|\geqslant 5$. Denote $\Sigma=\{p_1,\dots,p_k\}$ and assume that $h(p_i)=p_{f(i)}$ for $i=1,2,3,4,5$ and a certain (we may supposed that fixed) injective function $f:\{1,2,3,4,5\}\rightarrow \{1,2,\dots,k\}$. Consider two cases.
1) Assume that some $f(i)$ is greater than 5, say, $f(5)\notin \{1,2,3,4,5\}$. Then if we fix all points except $p_{f(5)}$, $h$ is uniquely determined by $h(p_i)=p_{f(i)}$ for $i=1,2,3,4$, and the event $h(p_5)=p_{f(5)}$ generically does not hold.
2) $f$ is a bijection of the set $\{1,2,3,4,5\}$, but not identity permutation (if $f$ is identity, then $h$ is identity).
2.1) If $f$ has a fixed point, say $f(5)=5$, again $h$ is determined by $h(p_i)=p_{f(i)}$ for $i=1,2,3,4$ (and $h$ is not identity), and the event that $p_5$ is a fixed point of $h$ generically does not hold.
2.2) $f$ is a cycle of length 5 or a product of disjoint cycles of lengths 3 and 2. The 5-tuple $\{p_1,p_2,p_3,p_4,p_5\}$ is projectively equivalent to the 5-tuple $\{p_{f(1)},p_{f(2)},p_{f(3)},p_{f(4)}, p_{f(5)}\}$. Take your favourite projective invariant of a 5-tuple, like the cross ratio of the points $p_1,p_2,p_3p_4\cap p_1p_2,p_3p_5\cap p_1p_2$ and verify that generically it changes under the permutation $f$ (it suffices to find one specific 5-tuple of points for which it changes).