Subgraph isomorphisms: does large out-expansion imply large in-expansion?
Yes: if each vertex maps to at least $k$ vertices by a set $S$ of bijections of a finite set, then there are at least $k$ vertices that map to each vertex by these bijections.

In fact, the in-spectrum (the sequence of in-expansions) is a permutation of the out-spectrum (the sequence of out-expansions).  Note that $f$ is a bijection iff $f^{-1}$ is a bijection.  Hence one can define a bijection between $S$ and $S^{-1} = \{f^{-1}\mid f\in S\}\ $ so the spectra are then permutations of each other.



An aside: this doesn't necessarily hold for bijections of infinite sets.  Let $f_i$ be the map $1\mapsto i, 2\mapsto 1, 3\mapsto 2, \dots, i \mapsto i-1, i+1\mapsto i+1,\dots$, and take the collection of all these maps.  Then the out-spectrum is $(\omega,2,2,\dots)$ while the in-spectrum is $(2,2,2,\dots)$.