Let 8 points define 10 or more similar quadrilaterals with no self-mapping. No self-mapping means that a square counts just once instead of 8 times. What is the maximal number of differently sized similar quadrilaterals in a solution? I think the answer is 5 different sizes.
$\{ \\ \ \ \ \ \{7,3,4,2\}, \{1,2,8,3\}, \\ \ \ \ \ \{2,3,1,4\}, \{3,2,7,8\}, \\ \ \ \ \ \{3,4,2,5\}, \{2,8,3,6\},\\ \ \ \ \ \{3,1,4,6\}, \{2,7,8,5\},\\ \ \ \ \ \{4,2,5,6\}, \{8,3,6,5\} \\ \} $
For the points, pick the positive complex root of $x^3-x-1=0$. Powers -1 to 3 give points 1 to 5. Point 6 is the origin. The small numbers give the distance in terms of powers of the square root of the plastic constant.
I know there are other solutions based on adding/using points to regular n-gons. I think an asymmetric quadrilateral from the octagon gives the most similar quadrilaterals, 16 with 8 points. What is the maximum number of similar quadrilaterals for $n$ points? I think the solution will initially be $2 n$. At 16 points, this method gives 38 similar polygons, shown below with lines and colored splines. I don't know if 38 similar polygons from 16 points is maximal.
Here is a principle yielding an easy counting method.
Let $P=-0.662359+0.56228i$ (one of the roots of $x^3-x-1=0$).
As I understand it, your points are, in the order $1,2... 8$:
$$P^{-1}, \ \ 1, \ \ P, \ \ P^2, \ \ P^3=P+1, \ \ 0, \ \ 1+P^{-2}, \ \ P^{-2}$$
(Edit: the two last ones have been corrected)
With these notations, it is rather easy to look for similar quadrilaterals $Q_1$ and $Q_2$. For example these ones:
$$P*\underbrace{[P^{-2},P^{-1},1,P]}_{Q_1}=\underbrace{[P^{-1},1,P,P^2]}_{Q_2}$$
by the very fact that multiplying by a complex number amounts to a similitude.
In a reciprocal way, it looks a reasonable assumption that all similitudes are obtained in this way.
Remark: A special category of quadrilaterals are those with (invariant) point $0$ (origin). They will provide similar quadrilaterals by looking for similar triangles (with the other vertices). For example:
$$P*\underbrace{[P^{-2},P^{-1},1,0]}_{Q_1}=\underbrace{[P^{-1},1,P,0]}_{Q_2}$$