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.



Alexandrov's rigidity in higher dimensions
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 $(x\_1,x\_2,x\_3,x\_4)=(1,2,1,2)$.


