Wrong:
"since infrared waves have a shorter wavelength"
Infrared has longer wavelength than visible and visible longer wavelength than ultraviolet .
White is a term for visible light mixed wavelengths. In the plot you can see that almost half of the sun's radiated energy arrives as visible light. The white buildings reflect this visible light which otherwise,impinging on the surfaces would be absorbed and turned into infrared by the interactions, adding to the arriving infrared.
What is absorbed and what is reflected depends on the chemical bonds of the surfaces, whether the incoming radiation can excite molecular states of the materials. Infrared is in frequencies/wavelengths of the black body radiation of bodies in the temperature ranges comfortable for the human body, so they easily raise the vibrational and rotational levels of solids and liquids and the kinetic energy of gasses.
37C curve seen here practically all in infrared, and lower temperatures more so.
Thus white paint will not reflect infrared as efficiently as visible, a large part of infrared will be absorbed as also some part of visible will scatter at the surface and degrade to infrared. Infrared can be reflected by metal mirrors, from the collective fields in metals . If you put aluminum foil in front of a heater you are sheltered from most of the heat which is reflected, but some of it is absorbed as can be seen by touching the foil.
If I expose an object to EM radiation only from the infrared spectrum, will it only reflect back infrared?
Yes, but most of it will be absorbed ( except by mirror metal surfaces) because the materials have the receptors for these wavelengths. This is due to the fact that larger wavelengths have photons with less energy which cannot excite higher energy levels.The energy of the photons goes as h*c/lamda where h is plancks constant, lamda the wavelength and c the velocity of light.
Is this true for other types of EM radiation?
No.Visible and ultraviolet by scatterings degrade their energy down to infrared frequencies, depending on the material.
Is it possible to make an object that looks white and absorbs a lot of infrared radiation?
Usually most of the infrared will be absorbed except by mirror metal surfaces.
If an object reflects most of the EM radiation that it receives of a particular wavelength λ, will it also reflect most of the radiation it receives of wavelengths less than λ (and absorb most of the radiation of wavelengths larger than λ)? Is this why objects that reflect most visible light (and hence look white) also reflect most infrared radiation (since infrared waves have a shorter wavelength)?
There is no such rule. It depends on the material and its chemical bonds.
You can't argue with a black object and a white object alone, as I think you partially understand in trying to build your thought experiment. You need a little bit more to define things properly. See whether the following helps.
Imagine a black object at a temperature $T_0$ and a white object also at $T_0$ inside a perfectly isolating box full of blackbody radiation at some higher temperature $T_1>T_0$ (i.e. without the black and white objects, this radiation is in thermodynamic equilibrium).
To understand exactly what would happen, you would have to describe the "colour" of your objects with emissivity curves that show emissivity as a detailed function of frequency. So your "black" and "white" would need to be defined in much more detail. You would also have to define the surface areas of the two objects and what they are made of (i.e. define their heat capacities). But all of this only effects the dynamics of how the system reaches its final state, i.e. these details only influence how the system evolves. What it evolves to is the same no matter what the details: the box would end up with everything at the same temperature such that the total system energy is, naturally, what it was at the beginning of the thought experiment. "Blacker" as opposed to "Whiter in this context roughly means "able to interact, per unit surface area, with radiation more swiftly": the blacker object's temperature will converge to that of the radiation more swiftly than does that of the whiter object, but asymptotically the white object "catches up". Blacker objects absorb more of their incident radiation its true, but they also emit more powerfully than a whiter object at the same temperature. The one concept emissivity describes the transfer in both directions. Think of emissivity as being a fractional factor applied to the Stefan-Boltzmann constant for the surface as well as being the fraction of incident light absorbed by the surface relative to a perfect blackbody radiator.
This description is altogether analogous to that of the situation where $T_0<T_1$. Begin with $T_0=T_1$, and you've got thermodynamic equilibrium from the beginning. Nothing happens, of course.
Maybe the following will help thinking about what is a really quite a complex question: it would be a fantastic last question for an undergrad thermodynamics exam BTW: You can abstract detail away by saying lets define object $A$ to be blacker than object $B$ if, when both objects are made of the same material, are the same size and shape, the temperature of $A$ converges to the final thermodynamic equilibrium temperature more swiftly than that of $B$ when they are both compared in the box-radiation-object thought experiment above.
Thinking about this now, I am not sure whether the above definition would hold for every beginning temperature of the radiation. Maybe there are pairs of surfaces whose relative blackness is different at different beginning temperatures such that $A$ is blacker than $B$ with some beginning temperature whilst the order swaps at a different beginning temperature. I think it is unlikely, but that is probably a different question altogether.
By the way, which pub do you drink in? I might come along.
Afterword on a Heater's Colour:
You ask by implication what is the best colour to paint a heater. This is not a simple question and involves the dynamics of the heater system. It's really an engineering question. I suspect in general it is better for them to be blacker rather than whiter. Here's a glimpse of the kind of factors bearing on the situation.
If you can say a heater has a constant nett input of $P$ watts, then at steady state that's going to be its output to the room, altogether regardless of its colour. There may be a materials engineering implication here: if you paint the heater whiter, and if its dominant heat transfer to the room is by radiation (rather than by convection or conduction), then it has to raise itself to a higher temperature than it would were it blacker so as to radiate $P$ watts into the room. So its materials might not be as longlasting, and it might be more of a fire hazard than it would be were it blacker.
If the heater is the hot water kind, and again if radiative transfer is significant, then the heating system has to run hotter to output power at a given level if the heater is whiter. At a given flow rate and given temperature of heating water, the heat output of heater is lower if it is whiter. You're trying to design the heater to be an "anti-insulator": you want the heat to leak out of the flow circuit in at the heater, not through the lagging on the hot water pipes outside the building channelling the water from the boiler to the heaters. If the hot water pipes leak heat in the same room, then that's no problem.
Recall the quartic dependence of the Stefan Boltzmann law. At room temperatures with a low temperature heater (the hot water kind) $\sigma\,T^4$ is likely to be pretty small compared with other heat transfer mechanisms, in contrast to my idealised scenarios above. So the heater's colour is likely to be pretty irrelevant.
Best Answer
Even considering the same fabric with different colours it will depend a lot on optical properties of the dye which we cannot tell only by the colour that we can see. I would need to know "how much the silver paint would reflect and in what spectral range" and also "how is the absorption coefficient as a function of the wavelength (from near-UV, Visible to near-Infrared)". I believe that this is the most important point and not the colour that we perceive with our eyes.
Nevertheless, keeping things simple, I think that the reflective coating being outside or inside would make the same amount of radiation that passes through. The only difference is that putting the reflective layer inside you would be increasing the path length of the radiation, thus increasing the absorption, and therefore the temperature
EDIT:
In order to clarify my idea I added this picture. Dashed layer is reflective and the the grey layer is the normal dark layer. I would like to divide this discussion in two: "Radiation Protection" and "Heat Protection"
Regarding the radiation protection, I believe that it depends on how the the dark layer absorbs light in all spectral range of the sun light, i.e. absorption coefficient and thickness of the material+dye. It also depends on how the reflective layer reflects the light, i.e. reflectance. I believe that the radiation protection does not depend on the arrangement, so both schemes protect from radiation as good. (I am not considering far infrared radiation)
Regarding Heat Protection, the thought is the same, except we should consider that in the first case the length of the radiation in the dark layer is longer, thus absorbing more radiation thus heating more. Also, the protection from the heat will depend on the heat conductivity of the layers and surface morphology which will affect how well the layer will cool down.
I understand that having a reflective coating outside an umbrella would be unpleasant for other people. But it would be the best choice for heat protection. However, regarding the emergency heat blankets, I do not know why they also have a reflective layer on the inside.