Do all objects at the same temperature glow the same color? Does Kirchhoff's law for heat radiation imply that all objects at the same temperature will glow the same color?

In other words, if a piece of molten iron glows the same color as my body, which radiates the same color as the sun, therefore all three objects - molten iron, my body, and the sun - must be at the same temperature?
 A: 
If a piece of molten iron glows the same color as my body, which radiates the same color as the sun, therefore all three objects - molten iron, my body, and the sun- must be at the same temperature?

There is no one-to-one correspondence of color with temperature the way you envisage.
Paint is at room temperature, but can display any of the dominant colors of radiation from a heated object. Even your figure above shows this: is the computer screen surface at temperatures of thousand centigrade?
It is not simple. In this article the heat to energy density of matter  is shown using the black body radiation formula


The peak wavelength and total radiated amount vary with temperature according to Wien's displacement law. Although this shows relatively high temperatures, the same relationships hold true for any temperature down to absolute zero. Visible light is between 380 and 750 nm.

A: No. A classic example is the gas mantle, which glows white when most other materials would glow red. It works by using a material with an emissivity that is low at the red end of the spectrum, but high at the blue end.
A: Not really.
Every sample of material has an emission spectrum and corresponding absorption spectrum for any given temperature. The emission spectrum is exactly what it says on the tin: it is the spectrum of light emitted by the object when excited to do so, such as by being hot or, to say it another way, it is how susceptible the object is to emit light at each given frequency, which thus can vary depending on the frequency. The two spectra are complementary in that at any frequency an object is a good emitter, it is also a good absorber, and conversely: this follows from the fact of the time reversal symmetry of physics.
Planck's law applies to a "perfect black body". Such an object has a perfectly uniform emission and absorption spectrum, so it is equally responsive to light of all frequencies. Once heated, it gives off the characteristic glow color you describe. This is a spherical cow, just like a perfect vacuum, perfect conductor, rigid body, and so forth. For a realistic object, the actual emission will be Planck's law times the actual emission spectrum. However, for most materials the curve will still be close enough to a Planck curve that you will not easily be able to see much of a difference in the apparent glow, e.g. there is no known material, I believe, that can both be heated to the necessary temperature without being damaged in some way and also glows a green color, i.e. the emission spectrum is such that the reds are totally suppressed.
A: Kirchhoff's law only applies to a black body - an idealised object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Stars like the sun are a reasonably good approximation of a black body. It is unlikely that molten iron or your body would approximate a black body.
A: The glow (ie. the thermal radiation) of every material at the same temperature is the same color, but there are materials that emit light besides their thermal emission. This can be observed, for example, in ion-colored flames, where specific electron transitions are the cause of the light emission, or in computer monitors.
Besides emissivity, materials also reflect or transmit light - this reflection and transmission is often wavelength dependent and causes most of the colors in everyday objects.
A: Yes, all the bodies glow same color at given temoerature and wavelength at thermal equilibirium according to kirchoff's law. What difference is that intensity of different objects is different, not the shape of spectral curve. So a ceramic glow with same color as iron at same temperature. It is stated as,$$\frac{e_1}{a_1}=\frac{e_2}{a_2}=e_b$$ Where, $e_1$ and $e_2$ are emissivity of a body 1 and 2, $a_1$ and $a_2$ are absorotivity of same bodies. $e_b$ is emissivity of black body whose absorptivity is one, absorb all radiation at all wavelength and act as universal function from which emissivity of other bodies can be known.
