Given two isomorphic convex uniform polytopes in $R^n$, is there a way to embed them in a higher dimension so the vertices are all unit distance apart $P$ and $Q$ are two isomorphic convex uniform polytopes in $\mathbb{R}^n$. Is there a space $\mathbb{R}^m$, where $n \lt m$, in which we can embed $P$ and $Q$, such that for each vertex $p\in P$ and $q \in Q$, we have $d(p,~q) = 1$? We assume the embedding is isometric over each polytope individually.
As an example, two simplices with $n$ vertices unit distance apart in $\mathbb{R}^{n-1}$ can be embedded into the simplex with $2n$ vertices in $\mathbb{R}^{2n-1}$. This can be done in such a way that every pair of vertices between the two simplices is also unit distance apart.
Does this work for other convex uniform polytopes or are simplices a special case?
 A: Provided $P$ and $Q$ both are circumscribable, that is, both have a well-defined center point and their vertex distances from that center point is the same for any vertex each, then we have well-defined circumradii $r(P)$ and $r(Q)$ respectively.
A) Outlining the details of the comment:
Consider we'll embed $P$ into the double dimensional space as $(P, 0)$ and $Q$ likewise therein as $(0, Q)$, using the respective centerpoints as the common origin, then obviously all connecting (lacing) edges between pairs of vertices $p\in P$ and $q\in Q$ obviously are of size $d(p, q) = \sqrt{r^2(P) + r^2(Q)}$ by means of the law of Pythagoras, which thus is fully independent of the respective choice of vertices.
That is, your quest would be met only if additionally that radicant would come out to be unity.
B) Outlining the details of your own example:
Consider instead we'll embed $P$ into the double-plus-one dimensional space $(P, 0, 0)$ and $Q$ likewise therein as $(0, Q, h)$ for some $h>0$ shifted orthogonal copy, then we get by the same rule $d(p, q) = \sqrt{r^2(P) + r^2(Q) + h^2}$ instead, once again independently of the vertex choices.
And again we conclude that your quest would be met only if additionally that radicant would come out to be unity. For sure, that additional shift makes this constraint even more restrictive. And, obviously, solution A) was nothing but the special case of $h=0$ wrt. B).
Nota bene:
The construction according to A) is known to be the tegum product, while that according to B) is known to be the pyramid product of the respective polytopes $P$ and $Q$. Cf. https://bendwavy.org/klitzing/explain/product.htm
--- rk
A: This is not possible if your polytope is too large.
Suppose that $P$ has vertices $p_1,p_2\in P$ of Euclidean distance $d(p_1,p_2)=3$.
Suppose further that $P$ and $Q$ are embedded in $\Bbb R^m,m>n$ as desired, and that $q\in Q$. Then
$$3=d(p_1,p_2) \le d(p_1,q)+d(q,p_2) =2.$$
This is a contradiction.
It would still be interesting to see when this is possible for polytopes that fit in a unit sphere.
