Preservation of a Cartesian structure for a permutation group The notion of a Cartesian structure for a permutation group appears in this presentation on slide 10 (or 45). What I do not understand is what should it mean that a permutation group preserve such a struture?
To cite the relevant definitions:

A Cartesian structure on $\Omega$ is an identification of $\Omega$ with $A^d$, where $A$ is some set. We can regard $A$ an an "alphabet", and $A^d$ as the set of all words of length $d$ over the alphabet $A$. Then $A^d$ is a metric space, with the Hamming metric [...]: the distance between two words is the number of positions in which they differ.
  A Cartesian structure is non-trivial if $|A|>1$ and $d > 1$. Let $G$ be a primitive permutation group on $\Omega$. We say that $G$ is basic if it preserves no non-trivial Cartesian structure on $\Omega$.

What should it mean that a Cartesian structure is preserved? If $\Omega = A^d$ for some $|A| > 1$, $d > 1$ then tupels (or words) are mapped to tupels (or words), so what is (or is not) preserved here?
 A: It means that there is no injective homomorphism $G\to\operatorname{Isom}(\Omega)$.
That is, look at all the possible actions of $G$ on $\Omega$. If there is some action such that the Hamming distance between two words, $\alpha$ and $\beta$, is equal to the Hamming distance between $\alpha^g$ and $\beta^g$ for every element $g\in G$, then we say that $G$ preserves the Cartesian structure on $\Omega$.
To give an example, consider a set $A$ and identify $\Omega$ with $A^1$ — i.e. all words of length one. Then the Hamming distance is:
$$H_d(a,b) = 
\begin{cases}
0 & a=b \\
1 & a\ne b
\end{cases}$$
Thus any permutation group $G$ is not basic on $\Omega$. 
A: To say that $G$ preserves the structure $\Omega = A^d$ means that the action of $G$ on $\Omega$ is equivalent to a subgroup of the full automorphism group of the structure $A^d$.
This full automorphism group is a wreath product ${\rm Sym}(A) \wr {\rm Sym}(d)$. The base group is generated by those permutations that just permute the entries in each individual coordinate of $A^d$. This is the direct product of $d$ copies of ${\rm Sym}(A)$, one for each coordinate. A complement in the wreath product isomorphic to ${\rm Sym}(d)$ permutes the entries of each element of $A^d$. For example, if $d=2$, a generator of the complement would map each $(a,b)$ to $(b,a)$.
