Symmetric Bijective Algorithm for Integers
Use any 32-bit block cipher! By definition, a block cipher maps every possible input value in its range to a unique output value, in a reversible fashion, and by design, it's difficult to determine what any given value will map to without the key. Simply pick a key, keep it secret if security or obscurity is important, and use the cipher as your transformation.
For an extension of this idea to non-power-of-2 ranges, see my post on Secure Permutations with Block Ciphers.
Addressing your specific concerns:
- The algorithm is indeed symmetric. I'm not sure what you mean by "reverse the operation without a keypair". If you don't want to use a key, hardcode a randomly generated one and consider it part of the algorithm.
- Yup - by definition, a block cipher is bijective.
- Yup. It wouldn't be a good cipher if that were not the case.
I will try to explain my solution to this on a much simpler example, which then can be easily extended for your large one.
Say i have a 4 bit number. There are 16 distinct values. Look at it as if it was a four dimensional cube: 
(source: ams.org)
.
Every vertex represents one of those numbers, every bit represents one dimension. So its basicaly XYZW, where each of the dimensions can have only values 0 or 1. Now imagine you use a different order of dimensions. For example XZYW. Each of the vertices now changed its number!
You can do this for any number of dimensions, just permute those dimensions. If security is not your concern this could be a nice fast solution for you. On the other hand, i dont know if the output will be "obscure" enough for your needs and certainly after a large amount of mapping done, the mapping can be reversed (which may be an advantage or disadvantage, depending on your needs.)
The following paper gives you 4 or 5 mapping examples, giving you functions rather than building mapped sets: www.cs.auckland.ac.nz/~john-rugis/pdf/BijectiveMapping.pdf