So I have this side-view drawing.
Now I wonder, will such a multi-layer material have asymmetric heat conduction properties?
Namely, because of radiative conduction, reflective aluminum surface and absorbing paint surface, will heat conduction to the left of the drawing be higher than heat conduction to the right?
Will the effect be significant/noticeable at all?
Is it at all theoretically possible to construct material with such properties in the general case?
The second law of thermodynamics forbids materials that conduct better in one (forward) direction than the reverse direction - such a material placed between two containers at thermal equilibrium would drive the temperature away from equilibrium, decreasing the entropy of the whole system and paving the way for a perpetuum mobile...
Reflectance and absorbance of material is the same (at a given wavelength) which is why left-right and right-left processes will happen at the same rate.
You drawing will have anisotropy in the vertical versus horizontal directions - but not left-right versus right-left asymmetry. Sorry.
If we assume that both the aluminum foil and the paint have scalar heat conductivities then their composite will also be scalar, and thus symmetrical. Being both material either poly-crystalline or amorphous this is probably reasonable assumption.
@Floris suggested that I expand on this, but not being my area I can only summarize a few ideas from Truesdell's Rational Thermodynamics, chapter 7.
It is an old problem going back to Stokes (1851) whether the heat conductivity tensor $\mathbf{K}$, that is the tensor in the linear Fourier law between temperature gradient and heat flux written as $ \vec q = \mathbf{K} \nabla T $ for crystalline materials is always symmetric or not. Assume the simplest case of $\mathbf{K}$ independent of the temperature and deformation, then one gets from the 1st law $\rho c \partial_t{T} = \mathbf{K_{+}} \nabla^2{T}$ and from the 2nd law that $\mathbf{K_+}$ is positive definite, where $\mathbf{K_{\pm}} = \frac {1}{2} (\mathbf {K}\pm \mathbf{K^T})$. Notice that in the 1st law only the symmetric part of the tensor shows up, hence the natural inclination to ignore the skew part.
Stokes proved that for heat conduction there are 13 types of crystals and for seven of those types $\mathbf{K_-} = 0$, $\mathbf{K}$ symmetrical. Out of the 32 optical crystal classes 19 are forced to have symmetric conductivity tensor. In only two triclinic classes the skew part $\mathbf{K_-}$ may seem to have any value, and in the other cases at least one component will always vanish. Stokes conjectured that $\mathbf{K}$ mus always be symmetrical but he could not prove so. In the ensuing decades may experiments were conducted to verify it either theoretically or experimentally (Voigt, Curie, Soret, etc.) Gurtin (1969) supposedly proved that a rigid heat conductor cannot be thermally stable unless its $\mathbf{K}$ is also symmetric but these are nontrivial results.
It's tempting to think that heat will flow more readily across a cavity from a high emissivity surface to a low emissivity surface than in the opposite direction when the surface temperatures are interchanged, but this is fallacious. The reason is that repeated inter-reflection between the surfaces restores symmetry.
Call the radiant flux across the cavity from left to right $W_{LR}$ and the radiant flux from right to left $W_{RL}$, after all inter-reflection has been accounted for. Then if the left and right surfaces have emissivities $\epsilon_L$ and $\epsilon_R$ and absolute temperatures $T_L$, and $T_R$, respectively, the radiant balances at the two surfaces can be expressed as
Left surface: $W_{LR} = \epsilon_L \sigma T_L^4 + (1- \epsilon_L) W_{RL}$
Right surface: $W_{RL} = \epsilon_R \sigma T_R^4 + (1- \epsilon_R) W_{LR}$
On solving for $W_{RL}$ and $W_{RL}$ the net radiant heat flux from left to right is found to be
$$ W_{LR} - W_{RL} = \frac{\sigma (T_L^4 - T_R^4)}{\frac{1}{\epsilon_L} + \frac{1}{\epsilon_R} -1 } $$ which simply reverses sign when $T_L$ and $T_R$ are interchanged.
