My question relates to
Landau & Lifshitz, Classical Theory of Field, Chapter 2: Relativistic Mechanics, Paragraph 8: The principle of least action.
As stated there, to determine the action integral for a free material particle, we note that this integral must not depend on our choice of reference system, that is, must be invariant under Lorenz transformations. Then it follows that it must depend on a scalar. Furthermore, it is clear that the integrand must be a differential of first order. But the only scalar of this kind that one can construct for a free particle is the interval $ds$, or $ads$, where a is some constant. So for a free particle the action must have the form $$ S = -a\int_a^b ds\,. $$ where $\int_a^b$ is an integral along the world line of a particle between the two particular events of the arrival of the particle at the initial position and at the final position at definite times $t1$ and $t2$, i.e. between two given world points; and $a$ is some constant characterizing the particle.
For me this statements are inaccurate. (Maybe it's because I have few knowledge from maths and physics. However.)
Why should the action be invariant under Lorentz transformations? Is this a postulate or it's known from experiments. If this invariance follows from special theory of relativity, than how? Why the action should have the same value in all inertial frames? The analogue of the action in non-relativistic Newtonian mechanics is not invariant under Galilean transformations, if I am not mistaken. See e.g. this Phys.SE post.
It is stated "But the only scalar of this kind that one can construct for a free particle is the interval." Why? Can't the Lagrangian be for example $$ S = -a\int_a^b x^i x_i ds\,, $$ which is also invariant.
The derivation of the Lagrangian for a non-relativistic free point particle was more detailed I think. See e.g. this Phys.SE post. Does the relativistic case need some more detalization?