Generating circular shifts / reduced Latin Squares in Python

You can use collections.deque:

from collections import deque

g = deque([1, 2, 3, 4])

for i in range(len(g)):
    print list(g) #or do anything with permutation
    g.rotate(1) #for right rotation
    #or g.rotate(-1) for left rotation

It prints:

 [1, 2, 3, 4]
 [4, 1, 2, 3]
 [3, 4, 1, 2]
 [2, 3, 4, 1]

To change it for left rotation just replace g.rotate(1) with g.rotate(-1).

variation on slicing "conservation law" a = a[:i] + a[i:]

ns = list(range(5))
ns
Out[34]: [0, 1, 2, 3, 4]

[ns[i:] + ns[:i] for i in range(len(ns))]
Out[36]: 
[[0, 1, 2, 3, 4],
 [1, 2, 3, 4, 0],
 [2, 3, 4, 0, 1],
 [3, 4, 0, 1, 2],
 [4, 0, 1, 2, 3]]


[ns[-i:] + ns[:-i] for i in range(len(ns))]
Out[38]: 
[[0, 1, 2, 3, 4],
 [4, 0, 1, 2, 3],
 [3, 4, 0, 1, 2],
 [2, 3, 4, 0, 1],
 [1, 2, 3, 4, 0]]

For the first part, the most concise way probably is

a = [1, 2, 3, 4]
n = len(a)
[[a[i - j] for i in range(n)] for j in range(n)]
# [[1, 2, 3, 4], [4, 1, 2, 3], [3, 4, 1, 2], [2, 3, 4, 1]]

and for the second part

[[a[i - j] for i in range(n)] for j in range(n, 0, -1)]
# [[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]

These should also be much more efficient than your code, though I did not do any timings.

more_itertools is a third-party library that offers a tool for cyclic permutations:

import more_itertools as mit


mit.circular_shifts(range(1, 5))
# [(1, 2, 3, 4), (2, 3, 4, 1), (3, 4, 1, 2), (4, 1, 2, 3)]

See also Wikipedia:

A circular shift is a special kind of cyclic permutation, which in turn is a special kind of permutation.