When we talk about kinetic energy of a body, it is actually the combined kinetic energies of all the atoms in that body. Now suppose a body is at rest. So its atoms would have some internal random motion and thus we can measure its temperature. But suppose it starts moving with a constant velocity $v$, so it will have kinetic energy $\frac{1}{2} mv^2$. So the kinetic energy of each atom is increased. So this means that the body will becomes hotter than earlier. But from the frame of body itself, it was at rest and thus there should be no change in its temperature although the temperature of the atoms of the ground should change.
How is this even possible? Why is temperature of the same body different when seen from two different inertial frames? Does this mean that it is not compulsory that two inertial frames are equivalent?
Note : I tried to search for a similar question on this site and I found this question and the selected answer literally seems contradicting to me and my ideas. The answer talks about all three possibilities and all the possibilities are based on pure mathematics (I guess), so I would like to know
why is there even a possibility of the temperature of a moving body to fall down or remain constant? Shouldn't it increase?
Or if I am wrong somewhere please correct me.
Please try to give a purely physical reason. It would be more appreciated since I am just a high schooler and don't understand those transformational mathematics.
Edit: I don't know why is the question closed. The question linked (as a dupe of ) , gives a wrong information that since pressure is invariant that's why temperature is invariant too. But this implies that pressure is more fundamental than temperature which is not true. So it will be helpful if the question is reopened and someone provides a proper reason for temperature invariance.
Edit 2 : In thermodynamics we can change the internal energy of a system (and thus change the temperature) by doing work on it . Then why can't the work of friction change our body's temperature ?