Polya's theorem to number of $d$-tuples of integers modulo n The problem is to count the number of $d$-tuples of the integers mod $n$ where two points are considered equal if the coordinates of one are the inversions of the other. 
More formally, we have $G=\mathbb{Z}_2$ acts on $\mathbb{Z}_n$ by $(0,m)\mapsto m$ and $(1,m)\mapsto -m \pmod n$ and this induces an action on $X=\mathbb{Z}_n^d$. Using Burnside's lemma, I have calculated that,
$$|X/G|=\frac12[n^d+2^{d(n+1 \pmod 2)}]$$
and one can check $n=12$, $d=1$, this counts the $7$ inversion classes (in fact, jotting out the classes for a few $n$, with $d=1$, one can obtain the result by observation but I wanted to practice these tools). 
Now I want to calculate the same result using Polya's theorem but I am having some trouble. I believe I have the solution for $d$-combinations and $d$-permutations, where $d\leq n$ and no repetitions are allowed. But I am unsure which weights to choose for $d$-tuples. Any help is appreciated.
 A: This is Power  Group Enumeration with the group $A$  acting on the
$n$  slots containing  one permutation,  namely the  identity and  the
group  $B$  acting  on  the values  (the  $d$-tuples)  containing  two
permutations, the identity  and the permutation that  takes an element
of a $d$-tuple to its  inverse.  Unfortunately we only start profiting
from this  technique when the two  groups have cycle indices  of lower
count of partitions than  the order of the group. We  are lead back to
Burnside in the present case.
With  PGE  in  its  unrestricted  form we  place  complete  cycles  or
sequences thereof from the cycle index  of the group $B$ acting on the
values on the  cycles from the group  $A$ acting on the  slots. In the
present case  we have  one slot permutation,  the identity,  which can
only be  covered by fixed points  from the two permutations  acting on
the  $d$-tuples.  There  is  the identity,  which  contains $n$  fixed
points, for a  contribution of $n^d$ (we may place  any fixed point on
any fixed point of the slot permutation). And there is the permutation
that implements the  inversion, which has one fixed  point namely zero
when $n$ is odd and two fixed points when $n$ is even, namely zero and
$n/2.$ This  yields one choice  for the $d$  fixed points of  the slot
permutation when  $n$ is odd  and two choices when  $n$ is even  for a
contribution of $2^{d(n+1 \mod 2)}.$ This is the same as Burnside. The
standard procedure  would be to compute  the cycle index of  the power
group $B^A$ and apply PET, but here we only have two permutations.
Observe that the  choice of variables $X_0, X_1,  \ldots X_{n-1}$ does
not work here because the weight would not be identical on the orbits.
