Alexandrov's rigidity in higher dimensions If $\Phi_1,\Phi_2$ are convex polyhedra in $\mathbb{R}^3$ such that the sets of outer normals to facets coincide, but $\Phi_1$ is not a translate of $\Phi_2$,  then there exist two corresponding facets $F_1,F_2$ (with the same outer normal) such that one of them is a translate of a proper subset of another.
This is A. D. Alexandrov's theorem, which generalizes the theorem of H. Minkowski which assumes that the areas of corresponding facets are always equal.
If I remember well, then in dimensions greater than 3 this is no longer true (while Minkowski theorem holds true in any dimension.) The request is a reference to counterexamples.
 A: Here is a counterexample in dimension four.
Consider positive numbers $x_1,x_2,x_3,x_4\in\Bbb R$ with $x_1<x_2$ and $x_3<x_4$ and construct the two 4-orthotopes (cartesian products of intervals)
\begin{align}
O_1:=[0,x_1]\times [0,x_2]\times[0,x_3]\times[0,x_4]\\
O_2:=[0,x_2]\times [0,x_1]\times[0,x_4]\times[0,x_3]
\end{align}
A pair of parallel facets is defined by a 3-element subset $I=\{i_1,i_2,i_3\}\subset \{1,2,3,4\}$:
\begin{align}
F_1&:=[0,x_{i_1}]\times[0,x_{i_2}]\times[0,x_{i_3}] \subset O_1 \\
F_2&:=[0,x_{\sigma(i_1)}]\times[0,x_{\sigma(i_2)}]\times[0,x_{\sigma(i_3)}] \subset O_2
\end{align}
where $\sigma$ is the permutation $(12)(34)$. (strictly spoken, the inclusions are wrong, but I hope the idea is clear).
For $F_1$ to be (parallel to) a subset of $F_2$ it must hold $(*)\,x_i\le x_{\sigma(i)}$ for all $i\in I$.
But each 3-element subset $I\subset\{1,2,3,4\}$ contains either $\{1,2\}$ or $\{3,4\}$, and so $(*)$ cannot be satisfied.
The easiest example is probably $(x_1,x_2,x_3,x_4)=(1,2,1,2)$.
