Given a cycle $x \mapsto x^a$ with his starting point $x_1$. Can another starting point $x_2$ be transformed to generate the same cycle?

In analyzing cases like this, it is useful to look at the behaviors modulo $P$ and modulo $Q$ separately, and then see how they combine. I will add in the assumption that $P \ne Q$; you didn't explicitly say so; I believe it is reasonable that you assumed it.

When we look at the cycle structure modulo $P$, we first see that $s_0 \bmod P$ is a quadratic residue modulo $P$ (which is a mathy way of saying "it's a square"); since $a$ is a primitive root of $p$, then we see that there are three cycles:

• A cycle of length 1 (the value 0)

• Another cycle of length 1 (the value 1)

• A cycle of length $p-1$; this is because $a^i \bmod p-1$ are distinct values in the range $[0, p-2]$, and $s_0$ has order $p-1$ (in this group, quadratic residues other than 1 are always that order), and so $s_0^{a^i}$ are $p-1$ distinct values.

We get similar results for looking at the behavior modulo $Q$.

Given those basics, how do they combine?

Well, the cycle modulo $PQ$ repeats only when both the cycle modulo $P$ repeats and the cycle modulo $Q$ repeats; if the $P$-cycle is length $\alpha$ and the $Q$-cycle is length $\beta$, this combined cycle has length $\text{lcm}(\alpha, \beta)$.

This implies that any combination of the two large cycles will have length $\text{lcm}(p-1,q-1)$ (which is a result you have already found). And, there are $\gcd(p-1,q-1)$ ways these two large cycles can combine.

We now consider a combination that includes a small cycle; we have two combinations with cycle length $p-1$, two combinations with cycle length $q-1$, and four combinations with cycle length 1 (which includes $s_0 = 0$ and $s_0 = 1$).

Hence, the total number of cycles is $\gcd(p-1, q-1) + 8$.

And, to answer your question:

Is there any way to transform $x_2$ so it will produce the same cyclic sequence as $x_1$ does?

Well, for any integer $\beta$, we have $s_{i+1}^\beta = (s_i^\beta)^a$. That is, if we take every element of a cycle, and raise it to the power $\beta$, we still have a cycle.

So, if we have the value $\beta$ for which $x_2^\beta = x_1$, then that gives us a way to transform one cycle to another.

It turns out that there is always such a $\beta$ if the two cycles are both large (that is, of length $\text{lcm}(p-1, q-1)$). For the degenerate cycles (all the others), there won't be - however, I don't consider that case that interesting...

Of course, finding such a $\beta$ given $x_1, x_2$ is a nontrivial problem if $P, Q$ are large...