What are the all equivalence classes that exist for this relation? Let $f$ be a function that associates a set of $n$ points $P\subset\mathbb{R}^2$ with a graph $G$ in the following way:
If $S$ is the set of all line segments between each possible pair of points in $P$, then $G$ is the graph whose nodes represent the line segments in $S$ and whose edges represent their intersections. In other words, two line segments in $S$ are adjacent in graph $G$ if they are intersecting. Also assume that no subset of $P$ with cardinality greater than 2 is colinear.
Then let $\sim$ be a relation on $\mathcal{P}(\mathbb{R}^2)$ such that $P\sim Q$ if and only if the graphs $f(P)$ and $f(Q)$ are isomorphic (where $\mathcal{P}$ denotes the power set).
Then, for any given value of $n$, how can I identify each (or at least find the number of), equivalence classes for $\sim$?

It is at least intuitive for me that, and I am fairly certain that, when $P$ can be taken as the vertices of a convex polygon, this should represent a unique equivalence class.
However, I am unsure of where to go from there.
 A: We begin by counting the nodes of the intersection graph of open line segments between all pairs of points $u,v \in P$.  Since there are $n = |P|$ (non-collinear) points in the plane, there are $\binom{n}{2}$ open line segments that form the nodes of graph $\mathscr G$.
It is easy to give a sharp upper bound on the number of edges in $\mathscr G$.  Two nodes are connected by an edge if and only if their corresponding open line segments intersect, and this occurs if and only if their endpoints form a convex quadrilateral when taken in alternating order:

(Credit Gareth Rees for image at StackOverflow)
Since each set of four points in $P$ produces an edge in $\mathscr G$ (consisting of two intersecting diagonals) if and only if the corresponding quadrilateral is convex, graph $\mathscr G$ has at most $\binom{n}{4}$ edges. That many occur exactly when all our potential quadrilaterals are convex.  A little thought reveals this to be so precisely when the points $P$ form a convex $n$-gon.
This leads nicely into a proof that two convex $n$-gons $P,P'$ will give isomorphic intersection graphs $\mathscr G, \mathscr G'$.  Order consecutively the points $u_1,\ldots,u_n$ around $P$ and the points $v_1,\ldots,v_n$ around $P'$.  The points correspondence between $u_i$ and $v_i$ induces a consistent correspondence between the open line segments of $P$ and of $P'$.  Thus we have a mapping of nodes in $\mathscr G$ to nodes in $\mathscr G'$.
Finally the consistent ordering of points around each convex polygon ensures that a convex quadrilateral, composed of alternating endpoints of two segments in $P$, corresponds to a convex quadrilateral whose diagonals are intersecting open segments in $P'$.  Therefore the edges of $G$ map to edges of $G'$ under the induced correspondence of nodes, and conversely.  The two intersection graphs are isomorphic.
The analysis of cases where points of $P$ lie strictly inside its convex hull seems a daunting project.  One possible approach is to consider nesting of convex hulls, or convex layers.  For now it seems feasible only to work through some instances with very modest numbers of these interior points, hoping to account for additional graph isomorphism classes.
