SHA-1 collision for strings of two symbols Is it possible to have strings of two (Unicode) symbols with equal SHA-1 hash? 

For example, smth like "ab", "ba".
No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.



Edit:

As fgrieu correctly explained in the comment: Because of the birthday attack we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1,111,998^2$.



2nd Edit:

There are a lot of different sources for the exact amount of Unicode characters / symbols there could be (see comments below), but most of them are around $1'100'000$ and are either way far too small to reach $2^{80}$.