Very cool idea. Imagine there exist two distinct operating sequences of equal length that turn 3 into the same number. Take the smallest such sequences, meaning they differ at the last operation. Assume the number is x(x+1)/2 now, the number on the first sequence before hand was x, so at most x/2 + 1 operations had happened. And the second sequence was increasing by for at least (x(x+1)/2 - x(x-1)/2)/2 = x/2 operations and to get to x(x-1)/2 it took some operations as well, so we easily get that the number of total operations was more than x/2 + 1. So we are done. (The answer is 2^2023)

Answer is $\boxed{|A_{2023}|=2^{2023}}$ Assume that distinct sequence of operations starting with $3$ lead us to the same number. And assume that the number that we can reach on $2$ different ways with minimum operation be $\frac {2k(2k+1)}{2}$ While doing the last operation, we should have $2k^2+k-2$ and $2k$. From starting on $\frac {(2k-1)2k}{2} = 2k^2-k$ until obtaining the $2k^2+k$ we have to do at least $k-1$ more operations. However we can't do that $k-1$ more operations since $2k$ will be $2$ after these operations if we do as so. Thus contradaction. Thus the operations we have made will always give us different numbers and conclusion follows.