How to represent a list as a cycle

Assuming that your lists contain no duplicates you can use this:

set = 
{{a, b, c, d}, {a, b, d, c}, {a, c, b, d}, {a, c, d, b}, {a, d, b, c}, 
 {a, d, c, b}, {b, a, c, d}, {b, a, d, c}, {b, c, a, d}, {b, c, d, a}, 
 {b, d, a, c}, {b, d, c, a}, {c, a, b, d}, {c, a, d, b}, {c, b, a, d}, 
 {c, b, d, a}, {c, d, a, b}, {c, d, b, a}, {d, a, b, c}, {d, a, c, b}, 
 {d, b, a, c}, {d, b, c, a}, {d, c, a, b}, {d, c, b, a}};

DeleteDuplicates[
  RotateLeft[#, Ordering[#,1] - 1] & /@ set
]

{{a, b, c, d}, {a, b, d, c}, {a, c, b, d}, {a, c, d, b}, {a, d, b, c}, {a, d, c, b}}

---

Extension to lists with duplicates

A pairwise comparison with SameTest will always be slower than placing elements into a canonical form and using Union or DeleteDuplicates with the default algorithms. To that end I propose this for lists that may have duplicates:

canonize[a_List] := 
  With[{X = # ~Extract~ Ordering[#, 1] &},
    RotateLeft[a, # - 1] & /@ Position[a, X @ a] // X
  ]

Example:

SeedRandom[1];

set =
  Join[
    Table[RandomSample[{a, b, c, d}, 4], {20}],
    RandomChoice[{a, b, c, d}, {5, 4}]
  ];

Union[canonize /@ set]

{{a, a, a, c}, {a, b, c, d}, {a, b, d, c}, {a, c, a, d},
 {a, c, b, d}, {a, c, d, b}, {a, d, b, b}, {a, d, b, c},
 {a, d, c, b}, {b, c, b, c}}

This is far faster than a pairwise compare:

rotatedQ[{x___}, {y___}] := (* example function provided by whuber *)
  Length[{x}] == Length[{y}] && MatchQ[{y, y}, {___, x, ___}];

big = RandomInteger[4, {5000, 5}];

Union[canonize /@ big] // Length // Timing

Union[big, SameTest -> rotatedQ] // Length // Timing

{0.047, 629}

{5.007, 629}

A solution using the built-in Cycles comparison:

numbering = MapIndexed[#1 -> #2[[1]] &, Sort[set[[1]]]];
DeleteDuplicates[set, Cycles[{#1}] == Cycles[{#2}] /. numbering &]