Sufficient condition to inscribe a parallelogram inside another one

Question

This question comes after another question I've asked Sufficient condition to inscribe a polygon inside another one.

Given two parallelograms $P$ and $P'$ with vertexes respectively $x_1,\dots,x_4$ and $x_1',\dots,x_4'$, the condition $|x_i'x_j'|\le |x_ix_j|$ for all $i,j=1,\dots,4$ (i.e. all sides and diagonals of $P'$ are not longer than the corresponding sides and diagonals of $P$) is enough to conclude $P'\subset P$ (i.e. $P'$ can be moved by isometries in such a way to be entirely contained inside $P$)?

In my previous question user Misha showed a counterexample which clearly fails in case $P$ and $P'$ are both parallelograms

Answer 1

We can construct a degenerate parallelogram with side lengths 10 and 20 and diagonal lengths 10 and 30. It cannot contain a unit square.

Answer 2

No, counter example:

The rhombus A'B'C'D' has the same side length as the unit square ABCD with. The diagonal of the rhombus is at most 2 =(2*side length), which is bigger than the diagonal of the square ($\sqrt{2}$). Thus the rhombus can never fit in the square.