For graphs $G$ and $F$, we say that $G$ is $F$-free if $F$ does not occur as an induced subgraph of $G$. This paper is concerned with the following question: Given an $F$-free graph $G$ having two vertices between which there exists at least one Hamiltonian path, how many Hamiltonian paths between these endpoints must exist (in terms of the order of $G$)? Our main result shows that there exists a sharp dichotomy. More precisely, we show that if $F$ is not an induced subgraph of $P_3+sP_1$ for any integer $s \geq 0$, then there exists an infinite family of $F$-free graphs having two vertices between which there exists a unique Hamiltonian path. On the other hand, we prove that if $F$ is an induced subgraph of $P_3+sP_1$ for some integer $s \geq 0$, then any $F$-free graph having two vertices between which there exists a Hamiltonian path contains exponentially many such paths between these two vertices. Our proofs use Ramsey-theoretic methods, a result on the existence of two vertices with low degree in graphs containing a unique Hamiltonian cycle, a path variant of Thomassen's red-independent weakly green-dominating sets, and a structural analysis of Hamiltonian paths in $P_3+sP_1$-free graphs. As an algorithmic consequence we obtain that for every fixed $s \geq 1$, given a Hamiltonian $sP_1$-free graph together with a Hamiltonian cycle, one can decide in linear time whether a second Hamiltonian cycle exists and construct one if it does.