Why is GCM not vulnerable at the birthday bound for 128 bit IV?
Actually, it turns out that GHASH on 128 bit values is a permutation (unless $H=0$; this happens with probability $2^{-128})$

GHASH of a 128 bit value $X$ simplifies to $\text{GHASH}_H(X) = H \times X$, where $H$ is a function of the key (that is, will always be the same for a specific key), and $\times$ is multiplication in the field $GF(2^{128})$.

This is a permutation, as $GF(2^{128})$ is a field, and multiplication in a field by a fixed nonzero value is always a permutation.



As for the 'collision' you mention in your comments, that actually can't happen.

If we have $H \times I_1 = 128$, that means that $H \ne 0$ and $I_1 = 128 \times H^{-1}$

If we have $H \times ((H \times I_2) \oplus 128) = 0$, that can happen only if $H = 0$ (which can't happen if the first equation holds) or $(H \times I_2) \oplus 128 = 0$, which holds only if $H \times I_2 = 128$, that is, if $I_2 = 128 \times H^{-1}$

Hence, for both to hold, we must have $I_1 = I_2$, and thus this isn't a 'collision'