Probability amplitudes in Feynman's path integral formalism

I spent the last 2 hours reading up on Feynman’s path integral formalism of quantum mechanics. Now, I have a few questions, which are rather very simple, but since unconventional (or maybe silly) are not present in any text that I could find online.

Define the propagator of a quantum system between two spacetime points ($x′$, $t′$) and ($x_{0}$, $t_{0}$) to be the probability transition amplitude between the wavefunction evaluated at those points.

$$
K(x’, t’ ;x_{0}, t_{0}) = langlepsi(x’, t’) | psi(x_{0}, t_{0})rangle.
$$

If the Hamiltonian carries no explicit time-dependence, we can relabel the first timevalue $t_{0} = 0$ and work only with elapsed time $t = t′ − t_{0}$. Now, according to Feynman’s formulation, we have:

$$
K(x’, t’ ;x_{0}) = A(t)sum_{alltext{ }trajectories}exp[i S[x(t)]]
$$
where $x(t)$ is a trajectory and $S[x(t)]$ is the classical action of that trajectory. Every possible path contributes with equal amplitude to the propagator, but with a phase related to the classical action. Summing over all possible trajectories, we arrive at the propagator. The normalization constant A(t) is independent of any individual path and therefore depends only on time.

Now, let $x_{cl}$ denote the classical trajectory of the particle. The classical trajectory is a unique trajectory given a particle under a potential in a given time. But quantum mechanically, the particle has a probability of being in multiple places after a given time.

To find the transition amplitude of a particular transition, say from $x_{0}$ to $x’$ we need to sum through all possible paths to get from $x_{0}$ to $x’$. The classical trajectory has the highest contribution to the above summation and hence reinforces the classical notion for the limit of ${hbarto0}$.

Now, isn’t the classical trajectory unique? So now if you want to find the transition amplitude from say $x_{0}$ to $x”$, we sum the contributions from all the paths along with their phase. Since the classical trajectory takes the particle from $x_{0}$ to $x’$ and not $x”$, we do not have a classical trajectory in the summation of the transition amplitude from $x_{0}$ to $x”$. But every text I read says that there is a term for the classical trajectory which dominates. But shouldn’t that be the case only for the transition amplitude $x_{0}$ to $x’$ (and not any other x) since $x_{0}$ to $x’$ is the $textit{only}$ classical trajectory assumed here? What am I getting wrong here?