This is A000940. The numbers grow fairly quickly, so for many applications, looking up a number in that list might be enough. I'll quote the terms up to the 20-gon:
3 1
4 2
5 4
6 12
7 39
8 202
9 1219
10 9468
11 83435
12 836017
13 9223092
14 111255228
15 1453132944
16 20433309147
17 307690667072
18 4940118795869
19 84241805734539
20 1520564059349452
I used the following naive python code to enumerate the first few items and identify the sequence:
import itertools
def f(n):
r = set()
for p in itertools.permutations(range(n)):
# cyclical shifts, correspond to a change in starting point:
s = set(p[i:] + p[:i] for i in range(n))
# include the reversed tuples, traverse path in opposite direction:
s |= set([tuple(reversed(i)) for i in s])
# modulo-add an integer to all elements, represents a rotation:
s = set(tuple((i + k) % n for k in j) for i in range(n) for j in s)
# also modulo-negate the elements, represents a reflection:
s |= set(tuple(n - j - 1 for j in i) for i in s)
s = frozenset(s)
r.add(s)
return r
for n in range(1, 10):
print("{} {}".format(n, len(f(n))))
Hmmm. Come to think of it, that Maple code given there could indeed be read as a formula:
$$f(n)=
\frac1{4n^2}\left(
\sum_{d\mid n}\left(\left(\varphi\left(\frac nd\right)\right)^2
\cdot d!\cdot\left(\frac nd\right)^d\right)
+\begin{cases}
2^{\frac{n-1}2}\cdot n^2\cdot\left(\frac{n-1}2\right)!
& \text{for $n$ odd} \\
2^{\frac{n}2}\cdot\frac{n(n+6)}4\cdot\left(\frac{n}2\right)!
& \text{for $n$ even}
\end{cases}
\right)$$