On a totally bounded metric space every non-contractive self-map is an isometry. For quasi-metric spaces, taken $T_1$ throughout ($d(x,y) = 0$ implies $x = y$), this fails: Zavarzina gave a hereditarily precompact counterexample and asked whether bijectivity restores rigidity; a recent example, the toll spiral, shows that it does not. We prove that compactness does: every compact quasi-metric space is expand-contract plastic, that is, each of its bijective non-contractive self-maps is an isometry. Countable compactness suffices, as does compactness of the conjugate topology. Bijectivity cannot be dropped: a compact, hereditarily precompact quasi-metric on the natural numbers makes the shift non-contractive and not an isometry. For arbitrary non-contractive self-maps we prove rigidity when the space is hereditarily precompact and its conjugate topology is sequentially compact, in particular when it is hereditarily precompact and Yoneda complete; neither hypothesis can be dropped. In both theorems the $T_1$ axiom can be weakened to $R_0$ ($d(x,y) = 0$ implies $d(y,x) = 0$), with hereditary precompactness read sequentially (every sequence has a left K-Cauchy subsequence), but not to $T_0$. We construct a $T_0$ space with all the remaining hypotheses of both theorems, Yoneda complete and compact both in its own topology and in the conjugate one, carrying a bijective non-contractive self-map that is not an isometry. Under $R_0$ the two-sided case reduces to the metric theorem, both topologies being sequentially compact exactly when the symmetrized distance is compact; the one-sided theorems do not reduce to it, and plasticity does not pass to the symmetrization even for compact spaces. The two theorems are machine-checked in Lean, and so are the counterexamples, apart from a few statements listed at the end.