If NF is consistent, then yes Con(NF) would be one of these statements. NF can interpret finite order arithmetic, so by that it would be subject to Godel incompleteness theorems. If Randall Holmes's [proof][1] of Con(NF) is correct, then NF is slightly stronger than finite order arithmetic, this means that all [strong axioms of infinity][2] are independent of it.

 


  [1]: https://math.boisestate.edu/~holmes/nfproof/newattempt.pdf
  [2]: https://en.wikipedia.org/wiki/New_Foundations#Strong_axioms_of_infinity