Shalom [edit: originally M. Burger] showed that the pair $(\mathrm{SL}_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$ has Relative Property (T) with respect to standard generating sets.

(The action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathbb{Z}^2$ is the usual one, i.e. the semidirect product can be thought of as a group of affine transformations $x \mapsto A x + b$ where $A \in \mathrm{SL}_2(\mathbb{Z})$ and $b \in \mathbb{Z}^2$.

If we reduce $\mathrm{mod}\ p$, we can think of this as giving an "efficient" way of generating the translations $x \mapsto x + b$ for $b \in F_p^2$.)

A one-dimensional variant in the finite case is whether there exist bounded size subsets $S_p \subset F_p^{\times} \ltimes F_p$ and $\delta > 0$ such that the relative Kazhdan constant:

$\kappa\ (F_p^{\times} \ltimes F_p, F_p, S_p) \ge \delta$

i.e. whether the pairs $(F_p^{\times} \ltimes F_p, F_p)$ can form a relative expander family.

An equivalent formulation: do there exist bounded size sets $S_p$ of affine transformations on $F_p$, such that no non-empty subset $U \subset F_p$, $|U| \leq p/2$ is *almost invariant* with respect to all of them, i.e.

$\neg \exists U: \forall s \in S: |s(U) \cap U| > \frac{99}{100} |U|$

I believe the answer is no if one uses standard "generating" sets (they needn't actually generate) such as $x \mapsto x + 1,\ x \mapsto ax$, even if $a$ is allowed to vary with $p$. This is very slightly surprising, as these do generate all translations "efficiently" in the weaker sense of logarithmic diameter.

Is there a good argument as to why this should fail in general? Or might there be cunning sets $S_p$ such that relative expansion occurs?