Say I have built a simple working AC 50 Hz transformer. It has copper coils at each end of a soft iron rod.
If I slip an aluminum tube over the soft iron rod, so the copper coils now are over the aluminum tube, can this still work as a transformer?
Would not the aluminum tube block the magnetic fields?
Aluminium cannot shield static magnetic fields, but as described by Lenz's Law, a changing magnetic field will generate eddy currents in the aluminium that will in turn generate magnetic fields that oppose and partially cancel out the initial magnetic field.
For the geometry described here, the eddy currents will circulate circumferentially around the aluminium tube in the opposite direction to the AC current in the copper coil. For a perfect conductor, these currents would completely cancel out the magnetic field inside the tube. For a real conductor such as aluminium with resistivity $\rho$ and magnetic permeability $\mu$, the strength of the interior AC magnetic field will not be zero but will instead fall off exponentially from the surface to the interior with the penetration scale given by the skin depth:
$$\delta=\sqrt{\frac{\rho}{\pi f \mu}}$$
The skin depth of aluminium at a frequency $f= 50$ Hz is about a centimetre, so a thin-walled tube will only slightly reduce the efficiency of your transformer, first by reducing the magnetic field reaching the soft iron core and secondly because of resistive heating losses due to the eddy currents flowing in the aluminium. If the thickness of the walls of the aluminium tube are comparable to the skin depth, however, so little magnetic field will penetrate to the soft iron rod that it will be irrelevant.
Note that solid soft iron is not a great choice for a transformer core. For soft iron with a resistivity $\sim 1\times 10^{-7}\,\mathrm{\Omega\,m}$ and permeability $\sim 0.006\,\mathrm{H/m}$, its skin depth is less than a millimetre. Because of this, magnet cores are usually made of thin laminated iron/steel (to break up the eddy currents) or ferrite. (which has very high electrical resistivity).
You may also want to look at the answers to