H-N theorem says what you write between parenthesis only.In Newtonian physics, yes. Not in relativity. That is what the Herglotz-Noether theorem says (it's the theorem that says that all possible Born rigid motions can be described with only three degrees of freedom). Motions that do not meet the conditions of the H-N theorem are not rigid when relativistic effects are taken into account, even though they might seem to be in the Newtonian approximation.
Nobody stated that, only that one has to stipulate it tacitly without justification within the theory,and the claim is made by Born and Einstein.Not if it's going to be used to justify the statement that it's impossible to have measuring rods and clocks in SR. Then it does need some discussion in order to refute that erroneous statement. See my next post.
This was already settled by Laue in 1911 when it showed that rigid bodies cannot exist in SR so it is not worth arguing. You can accesss the paper here googling:"On the discussion concerning rigid bodies in the theory or relativity" by von Laue.This is not correct.
Minkowski spacetime does have a metric tensor for sure, I used the word metric with the meaning it has for instance in "metric space". And of course Minkowski spacetime is defined as an affine space with a metric tensor.The lack of a positive definite metric does not mean Minkowski spacetime does not have a metric and is just an affine geometry.