Definition of inertial frame of reference: All bodies in inertial motion describe linear relations in space-time coordinates.
Definition of two bodies in same state of inertial motion: Relative to a third body or inertial coordinate system, they have the same velocity as a vector.
Definition of length of a material body in a uniform state of inertial motion: The difference in position of its two ends as measured simultaneously.
Thus we have for a body in a state of inertial motion described by velocity $$\vec{u}$$ in a certain inertial coordinate system, the following linear relationship for it's endpoints A and B:
$$\vec{x}_{A,1} - \vec{u} t_{A,1} = \vec{x}_{A,2} - \vec{u} t_{A,2} \\ \vec{x}_{B,3} - \vec{u} t_{B,3} = \vec{x}_{B,4} - \vec{u} t_{B,4} $$
So for any given time, t, we have
$$\vec{x}_A(t) = \vec{x}_{A,1} + \vec{u} \left( t - t_{A,1} \right) = \vec{x}_{A,2} + \vec{u} \left( t - t_{A,2} \right) \\ \vec{x}_B(t) = \vec{x}_{B,3} + \vec{u} \left( t - t_{B,3} \right) = \vec{x}_{B,4} + \vec{u} \left( t - t_{B,4} \right)$$
So the length measured between comoving endpoints A and B at any one time is:
$$ L_{AB} = \left| \vec{x}_B(t) - \vec{x}_A(t) \right| = \left| ( \vec{x}_{B,3} - \vec{x}_{A,1} ) + \vec{u} \left( t_{A,1} - t_{B,3} \right) \right| = \left| ( \vec{x}_{B,3} - \vec{u} t_{B,3} ) - ( \vec{x}_{A,1} - \vec{u} t_{A,1} ) \right| $$
We can simplify this somewhat if we use the property of real numbers that there is a zero, thus our coordinate system has an origin, even if this origin is purely a fiction of mathematical convenience.
Then $$\vec{x}_{A,1} - \vec{u} t_{A,1} = \vec{x}_{A,2} - \vec{u} t_{A,2} = \vec{x}_{A,0} \\ \vec{x}_{B,3} - \vec{u} t_{B,3} = \vec{x}_{B,4} - \vec{u} t_{B,4} = \vec{x}_{B,0}$$
For any given time, t, we have
$$\vec{x}_A(t) = \vec{x}_{A,0} + \vec{u} t \\ \vec{x}_B(t) = \vec{x}_{B,0} + \vec{u} t$$
And the length measured between comoving endpoints A and B at any one time is:
$$ L_{AB} = \left| \vec{x}_{B,0} - \vec{x}_{A,0} \right|$$
So the question of this thread is what is the length of the same material body in a coordinate system where it is not moving.
Newton and Special Relativity give different answers.
For special relativity we can transform space-time points $$( \vec{x}_{A,1} , t_{A,1} ) , \; ( \vec{x}_{A,2} , t_{A,2} ) , \; ( \vec{x}_{B,3} , t_{B,3} ) , \; ( \vec{x}_{B,4} , t_{B,4} ) $$ and recalculate the length in the new coordinates.
$$\vec{x}' = \vec{x} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{x} \cdot \vec{u} ) \vec{u} }{ u^2 } - \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \vec{u} t \\ t' = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \left( t - \frac{ \vec{x} \cdot \vec{u}}{c^2} \right) $$
Using $$\vec{x}_{A,1} = \vec{x}_{A,0} + \vec{u} t_{A,1}$$ we get this:
$$\vec{x}'_{A,1} = \vec{x}_{A,0} + \vec{u} t_{A,1} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{x}_{A,0} \cdot \vec{u} ) \vec{u} }{ u^2 } + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{u} t_{A,1} \cdot \vec{u} ) \vec{u} }{ u^2 } - \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \vec{u} t_{A,1} \\ t'_{A,1} = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \left( t_{A,1} - \frac{ \vec{x}_{A,0} \cdot \vec{u}}{c^2} - \frac{ \vec{u} t_{A,1} \cdot \vec{u}}{c^2} \right) $$
which simplifies to this:
$$\vec{x}'_{A,1} = \vec{x}_{A,0} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{x}_{A,0} \cdot \vec{u} ) \vec{u} }{ u^2 }
\\ t'_{A,1} = \sqrt{1 - \frac{u^2}{c^2}} t_{A,1} - \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ \vec{x}_{A,0} \cdot \vec{u}}{c^2} $$
Similarly, we have:
$$ \vec{x}'_{A,2} = \vec{x}_{A,0} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{x}_{A,0} \cdot \vec{u} ) \vec{u} }{ u^2 } \\ t'_{A,2} = \sqrt{1 - \frac{u^2}{c^2}} t_{A,2} - \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ \vec{x}_{A,0} \cdot \vec{u}}{c^2} \\ \vec{x}'_{B,3} = \vec{x}_{B,0} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{x}_{B,0} \cdot \vec{u} ) \vec{u} }{ u^2 } \\ t'_{B,3} = \sqrt{1 - \frac{u^2}{c^2}} t_{B,3} - \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ \vec{x}_{B,0} \cdot \vec{u}}{c^2} \\ \vec{x}'_{B,4} = \vec{x}_{B,0} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ ( \vec{x}_{B,0} \cdot \vec{u} ) \vec{u} }{ u^2 } \\ t'_{B,4} = \sqrt{1 - \frac{u^2}{c^2}} t_{B,4} - \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ \vec{x}_{B,0} \cdot \vec{u}}{c^2} $$
So $$\vec{x}'_{A,1} = \vec{x}'_{A,2} = \vec{x}'_{A,0}$$ and $$\vec{x}'_{B,3} = \vec{x}'_{B,4} = \vec{x}'_{B,0}$$
So $$ L'_{AB} = \left| \vec{x}_{B,0} - \vec{x}_{A,0} + \frac{1 - \sqrt{1 - \frac{u^2}{c^2}}}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ \left( \left( \vec{x}_{B,0} - \vec{x}_{A,0} \right) \cdot \vec{u} \right) \vec{u} }{ u^2 } \right|$$
So if $$\left( \vec{x}_{B,0} - \vec{x}_{A,0} \right) \cdot \vec{u} = 0$$ then the motion was perpendicular to the length and there is no change in the length measured.
But if the directions are parallel, the length measure in the coordinate system where the object is not moving is greater than in the coordinate system where the length was moving with velocity u.
$$ L'_{AB} = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} L_{AB} > L_{AB}$$.
So if you accept that the length of a moving object is a thing, then length contraction is real, not illusory, in special relativity.
Whatever.