Delete duplicates in a list, depending on the sequence of numbers

This should do the job:

DeleteDuplicates[list, SameQ[#1, #2] || SameQ[#1, Reverse@#2] &]

If your application allows for reversed elements to be interchangeable you can do this:

DeleteDuplicates[Sort[{#, Reverse@#}][[1]] & /@ list]

The advantage here is speed; this is perhaps two orders of magnitude faster on large sets than the methods already presented.

Update: in versions 10+, you can use DeleteDuplicatesBy:

list = {{1, 2, 3}, {3, 2, 1}, {2, 1, 3}, {1, 2, 3}};
DeleteDuplicatesBy[list, Union[{#, Reverse@#}]&]

{{1, 2, 3}, {2, 1, 3}}

Original answer:

You can use your condition directly

 DeleteDuplicates[list, (#1 == Reverse[#2] &)]

or use Union with SameTest

 Union[list, SameTest -> (#1 == Reverse[#2] &)]

Edit: For the new requirement, you need to Or the conditions for "sameness" (as in @acl's answer)

 DeleteDuplicates[list, (#1 == Reverse[#2] || #1 == #2 &)]
 Union[list, SameTest -> (#1 == Reverse[#2] || #1 == #2&)]