Suppose we have a graph G. Is it true that we can map its vertices to the plane such that when connecting neighboring vertices with segments, then any induced cycle of G that has length at least 4 will have two edges that intersect?
What if I want each induced cycle of length at least 4 to be in a non-convex position? (If we write a bigger number instead of 4, then this would be some kind of strengthening of the theorem of Erdos-Szekeres.)
What if I want k induced cycles that do not intersect (any other or themselves)?
I suspect that the answer to all these questions is negative, i.e. there is a graph that we cannot linearly map with ruining all its induced cycles. I would also be very interested in any related results.
*Update March 19. I realized that I can show that there is a dense graph whose embedding will have an induced $C_4$ in convex position (whose edges might intersect). For a proof sketch, see my answer.
*Update April 3. Now I realized that if the graph is sparse in the sense that its degeneracy is constant, then its chromatic number is also constant, in which case we can put the vertices that belong to the same color class close to each other, which would mean that it is not possible to have too many disjoint cycles. So a graph with a linear number of edges has a drawing with at most a constant number of disjoint cycles.
*Remaining Question. Is there a method that guarantees non-crossing cycle(s) and does not use a counting argument but rather some topology?