If you could hash $16^{64} = 2^{256}$ distinct inputs to a hash with a 256 bit output, then you would expect some collisions. (This is equivalent to you rehashing every possible hash in the domain if you hash the 16-byte representation of every non-negative integer < $2^{256}$)
How many collisions? First lets assume the output of a hash function is uniformly randomly distributed. The probability of 2 hash values being the same (being a collision) is $(1/2^{256}) = 2^{-256}$
We have $2^{256}$ outputs, so there are $\frac{2^{256}*(2^{256} - 1)}{2}$ pairs of output hashes. Each of these pairs has probability ${2^{-256}}$ of being the same. So the expected number of collisions is ${2^{-256}} * \frac{2^{256}*(2^{256} - 1)}{2} \approx 2^{255}$
So this implies that with 2^256 inputs, you would expect half of them to be collisions with each other, and therefore for any new input you hash, there is a 50% probability of it colliding with a hash you already know. However 2^256 is a massive number, and getting anywhere near that many hashes is computationally infeasible today.
--EDIT--
As pointed out by fgrieu, even though there are $2^{255}$ collisions, some of the collisions are on the same element, thus the amount of hash values reached is instead significantly less than $2^{255}$.
I have been unable to find a source or derivation of how many distinct collision hashes would be reached. (An equivalent question "what is the number of distinct shared birthdays in a given room for the birthday problem")