The larger the red shift between the source of light and the mirror the larger the energy transfer from the photon to the mirror, when observed from the source of light, since the momentum vector of the photon is reversed by the mirror (in the mirror's frame). So, for small v/c, the momentum transfer to the mirror is almost independent of velocity v, actually (presumably, without cross-checking) proportional to 1-v/c, but the transfer of kinetic energy from the photon to the probe increases with the velocity of the mirror (seen from the source of light, and for "small" velocities), and hence the energy loss of the photon. (Momentum of an object is mass times velocity, while kinetic energy is proportional to mass times velocity squared. A fixed momentum transfer results in a fixed Δv, but kinetic energy transfer to non-relativistic probe of mass m is ΔE=0.5m x ((v+Δv)²-v²)=0.5m x (2vΔv+(Δv)²), hence approximately proportional to v, for small Δv << v, and for v << c.)

Or, seen from the mirror's perspective, we get red-shifted light with increasing velocity. Only this red-shifted portion of the photon is reflected in the mirror's frame (and presumably red-shifted a second time, when observed from the source of light). Photon energy is proprtional to its wave length (

photon energy equals frequency times Planck's constant).

For larger velocites, it's probably more complicated, since the number of reflected photons may be proportional to 1-v/c, and we'll eventually run into

relativistic Dopple shift, with an additional Lorentz factor. For v = 0.2c, the Lorentz factor is about 1-0.02, much closer to 1 than 1-v/c, so neglectible for a draft back-of-an-envelope calculation.