British Flag theorem: Let $P$ be a point in the plane, let $ABCD$ be a rectangle in the plane then:
$$PA^2+PC^2=PB^2+PD^2$$
Generalization: Let $ABCD$ be a rectangle in a plane, Let $P$ be a point in the Euclidean three-space, then:
$$PA^2+PC^2=PB^2+PD^2$$
Define two directly similar cuboids: Let $ABCDEFGH$ and $A_1B_1C_1D_1E_1F_1G_1H_1$ be two directly similar cuboids if $ABCDEFGH$ and $A_1B_1C_1D_1E_1F_1G_1H_1$ are two cuboids and:
$$\frac{AB}{A_1B_1}=\frac{BC}{B_1C_1}=\frac{AE}{A_1E_1}$$
Example: Two cubes are two directly similar cuboids.
Generalization: Inspiration from British Flag theorem: Let $ABCDEFGH$ and $A_1B_1C_1D_1E_1F_1G_1H_1$ be two directly similar cuboids in Euclidean three-space, then:
$$AA_1^2+CC_1^2+FF_1^2+HH_1^2=BB_1^2+DD_1^2+EE_1^2+GG_1^2$$
My question: The result is holds in Euclidean space ?
The following does not prove (or disprove) your conjecture. Instead, it only shows that it is not a generalization of the British Flag theorem, because it holds true in the degenerate case of a single point and an arbitrary quadrilateral, while the British Flag theorem only holds for rectangles.
Consider a cuboid $ABCDEFGH$ with congruent faces $ABCD$ and $EFGH$ (which are not necessarily rectangles). Flatten this cuboid such that at the limit $ABCD \equiv EFGH$. Now shrink the directly similar cuboid $A_1B_1C_1D_1E_1F_1G_1H_1$ down to a single limit point $P$.
Then the relation:
$$AA_1^2+CC_1^2+FF_1^2+HH_1^2=BB_1^2+DD_1^2+EE_1^2+GG_1^2$$
becomes:
$$PA^2+PC^2+PB^2+PD^2=PB^2+PD^2+PA^2+PC^2$$
which is always true.
In contrast, the British Flag theorem only holds true for $ABCD$ rectangles, and a variation thereof exists for parallelograms, but none such for arbitrary quadrilaterals.