Convex polygons inclusion relation
Here's an argument that you need time quadratic in the number of polygons. More precisely,
you should not be able to find containing pairs among $n$ $k$-sided polygons in time $O(n^{2-\epsilon})$, for any $k=O(n^{\delta})$ and any $\epsilon,\delta&gt;0$ It's a reduction from the orthogonal vectors problem, the problem of finding two disjoint binary vectors among a set of $n$ $k$-dimensional binary vectors, for which the non-existence of an algorithm with time $O(n^{2-\epsilon})$ is a standard conjecture in fine-grained complexity (see e.g.  Abboud et al, "More consequences of falsifying SETH and the orthogonal vectors conjecture"). Here orthogonality is measured in $\mathbb{Z}^k$, i.e. it means that the vectors have disjoint subsets of nonzero coordinates.

Given a set of vectors among which we wish to find an orthogonal pair, consider a regular $k$-gon $P$, whose vertices correspond to the coefficients of our vectors. For each vector $x$ construct a polygon $p(x)$ as the convex hull of the $k$-gon vertices for which $x$ has a nonzero coefficient. To make sure that none of these polygons contains each other, shrink each $k$-gon by a factor $1-\gamma |x|$ for some very small number $\gamma$, so that when one vector is dominated by another its polygon is shrunk more.

Next, form two more regular $k$-gons $Q$ and $R$, concentric with $P$, where $Q$ is slightly smaller than $P$ (so that all its vertices are inside the shrunken copies of vertices of $P$) and $R$ is slightly bigger (both close enough that the construction below produces convex polygons). For each vector $x$, construct a second polygon $q(x)$, a convex $k$-gon that, on each ray from the origin through a vertex of $P$, $Q$, and $R$, chooses either a vertex of $Q$ or $R$: a vertex of $Q$ when the corresponding coefficient of $x$ is nonzero, and a vertex of $R$ when it is zero.
We can make sure that all vectors have at least one zero coefficient in order to ensure that no $q(x)$ is contained inside a $p(y)$, and we can do the same variable-shrinking trick to ensure that no two $q$'s are nested.

With this construction, the only remaining containments of polygons are that a polygon $q(x)$ contains another polygon $p(y)$. This happens if and only if $x$ and $y$ are orthogonal. So if we could solve your polygon question in sub-quadratic time, we could also solve the orthogonal vectors problem in sub-quadratic time, conjectured to be impossible.