Does a photon's wavelength (and energy) change when reflecting off a mirror?

The reflected photons are entirely identical to the incident photons except for the change in direction. The energy lost due to the pressure exerted on the mirror can be accounted for by comparing the number $N_i$ of incident photons to the number $N_r$ of reflected photons. I.e., $N_i > N_r$.

The answer is "almost no" - the wavelength of the photon is virtually unchanged (in the initial rest frame of the mirror, the "lab frame"). Because the mirror is much more "massive" than the photon, it serves as a "momentum sink" and picks up almost no energy.

The best way to develop an intuition for this is to consider a collision between two balls: one lighter (with mass $m$) and initially moving (at velocity $v_1$) and one more massive (with mass $M$) and initially at rest. After scattering, the lighter ball leaves the scene at velocity $v_3$ and the more massive ball leaves the scene at velocity $v_4$.

Set $v_2 = 0$ in the following worked-out example (see page 3):

https://web.archive.org/web/20181222165457/http://www.its.caltech.edu/~teinav/Lectures/Ph%201a/Lecture%207%20-%202017-10-19.pdf

We obtain $v_3 = frac{(M-m)v_1}{M+m}$ and $v_4 = frac{2mv_1}{M+m}$

In the limit that $M >> m$ the fraction of the initial kinetic energy picked up by the massive object goes to zero, but it acquires twice as much momentum (and in the opposite direction) as the lighter object initially had. Thusly can momentum be transferred but (almost) no energy.

NOTE - In the "center of mass frame" the wavelength will be completely identical, but I believe it is the "lab frame" that supplies the intuition you seek. In the center of mass frame the momentum just changes sign and your equation should really be $|p| = E/c$, which admits a sign change of $p$ while conserving $E$. This is why there is a $2$ in the equation for $v_4$ - flipping the sign of momentum imparts double the initial momentum to the mirror.

In the consideration of the photon behaving like a particle and reflecting from the surface of the mirror, the momentum is conserved and the wavelength does not change.

From conservation of momentum we know that the total momentum of the photon and the mirror is the same before and after the collision (considering the photon and mirror to be a closed system). In the simplest of considerations of refelection the quantum of the momentum of the photon is the same, but the direction is different. Resolving the momentum into two parts, one parallel to the surface of the mirror and one normal to the surface of the mirror then the momentum in the direction normal to the surface is reversed. To conserve momentum the mirror itself must have a balancing momentum equal and opposite to the change in momentum of the photom in the direction normal to the surface of the mirror. (You can see this effect in a rotating 'solar windmill' toy left in sunlight on a window sill - the reflective 'sails' reflect light and turn).

I stated above 'in the simplest of considerations' because the process of interaction may not be straight forward. There are different ways in which the photon and the mirror interact. For example Compton scattering may occur and that will change the wavelength of the light. (See https://en.wikipedia.org/wiki/Compton_scattering for more details on Compton Scattering).

So in a simple 'particle' model of reflection momemtum is conserved even though the direction of the photon changes.

Even if you assume infinite mass for the mirror, and also assume infinite rigidity for the mirror, an aluminium mirror will only reflect about 90% of the light due to I squared times R loss. this will radiate black body (approx.), in the microwave range, therefore less energy will be available to be reflected. Assuming one photon only, the wave function must collapse and two new photon's will be created. one at microwave freq., and one at slightly lower freq. than the incident photon.