 Solution

No: ordering the attractive potentials does not impose the proposed ordering of reflection probabilities. Take $k=\pi /(4a)$ and let $M$ denote the particle mass. Choose $V_2=0$ everywhere andThese satisfy every stated ordering and support condition. The incoming energy is ${\hslash }^2{k}^2/(2M)$; the interior wavenumber in the finite square well is $q=\sqrt{k^2-2M{V}_1/{\hslash }^2}=3k$.

Here is an explicit matching calculation. For a general well of width $L=2a$, take exterior waves $e^{ik(x+a)}+r{e}^{-ik(x+a)}$ on the left and $t{e}^{ik(x-a)}$ on the right. Shifting the incident phase in this way does not change the reflection probability. Continuity of the wavefunction and its first derivative gives the propagation relationInverting it and adding the two equations after dividing the derivative equation by $ik$ yieldsThe incident and reflected exterior waves have the same speed, so their flux ratio isFor the chosen parameters $qL=3\pi /2$ and $(q^2-k^2{)}^2/(4k^2{q}^2)=16/9$. ThusThe zero potential has no reflected component. This proves the failure with explicit probabilities. More generally the same expression vanishes at $qL\in \pi \mathbb{Z}$, explaining resonant transmission through a square well and why increasing attractive well depth does not monotonically control reflection.