"Under what circumstances does a Hausdorff topology [properly] refine a unique compact Hausdorff topology?"

If $X$ is locally compact (non-compact, Hausdorff), then $X$ refines infinitely many distinct compact Hausdorff topologies. However, there are also examples of (non-compact, Hausdorff) spaces $X$ that refine a unique compact Hausdorff topology.

Theorem: If $X$ is a non-compact, locally compact Hausdorff topology that refines at least one compact Hausdorff topology, then it refines at least $|X|$ compact Hausdorff topologies.

Proof: See the proof of Proposition 4.3 in this paper. The basic idea is to take a one-point compactification of $X$, then take the point at infinity and glue it back down onto any point of $X$. Now you have a topology that's compact, Hausdorff, and refined by $X$. Choosing different targets for the gluing will result in different topologies. QED. [Note: These topologies, while different, may be homeomorphic, for example if $X$ is the real line.]

In contrast, we have the following example. Let $I = [0,1]$ (the set, not the topological space). Let $\sigma$ denote the usual topology on $I$, and let $\langle \sigma,A \rangle$ denote the topology on $I$ with subbasis $\sigma \cup \{A\}$. Let $A = \{\frac{1}{n}:n \geq 1\}$. Then I claim that $\sigma$ is the only compact Hausdorff topology refining $\langle \sigma,A \rangle$.

To see this, notice that if $a > 0$ then $\sigma$ and $\langle \sigma,A \rangle$ agree on $[a,1]$. Furthermore, since $[a,1]$ is compact Hausdorff, passing to $\tau$ will not change its topology, since any strictly coarser topology on $[a,1]$ fails to be Hausdorff. Thus $\sigma$ and $\tau$ agree on every $[a,1]$, and hence on $(0,1]$.

Thus $\tau$ is a (Hausdorff) one-point compactification of $(0,1]$ (with the usual topology). But there's only one of those! So $\langle \sigma,A \rangle$ has only one compact Hausdorff refinement, namely $\sigma$.

"Under what circumstances does a Hausdorff topology [properly] refine a unique compact Hausdorff topology?"

If $X$ is locally compact (non-compact, Hausdorff), then $X$ refines infinitely many distinct compact Hausdorff topologies. However, there are also examples of (non-compact, Hausdorff) spaces $X$ that refine a unique compact Hausdorff topology.

Theorem: If $X$ is a non-compact, locally compact Hausdorff topology that refines at least one compact Hausdorff topology, then it refines at least $|X|$ compact Hausdorff topologies.

Proof: See the proof of Proposition 4.3 in this paper. The basic idea is to take a one-point compactification of $X$, then take the point at infinity and glue it back down onto any point of $X$. Now you have a topology that's compact, Hausdorff, and refined by $X$. Choosing different targets for the gluing will result in different topologies. QED. [Note: These topologies, while different, may be homeomorphic, for example if $X$ is the real line.]

In contrast, we have the following example. Let $I = [0,1]$ (the set, not the topological space). Let $\sigma$ denote the usual topology on $I$, and let $\langle \sigma,A \rangle$ denote the topology on $I$ with subbasis $\sigma \cup \{A\}$. Let $A = I \setminus \{\frac{1}{n}:n \geq 1\}$. Then I claim that $\sigma$ is the only compact Hausdorff topology refined by $\langle \sigma,A \rangle$.

Let $\tau$ be any compact Hausdorff topology that is refined by $\langle \sigma,A \rangle$.

Notice that if $a > 0$ then $\sigma$ and $\langle \sigma,A \rangle$ agree on $[a,1]$. Furthermore, since $[a,1]$ is compact Hausdorff, passing to $\tau$ will not change its topology, since any strictly coarser topology on $[a,1]$ fails to be Hausdorff. Thus $\sigma$ and $\tau$ agree on every $[a,1]$, and hence on $(0,1]$.

Thus $\tau$ is a (Hausdorff) one-point compactification of $(0,1]$ (with the usual topology). But there's only one of those! So $\langle \sigma,A \rangle$ is refined by only one compact Hausdorff topology, namely $\sigma$.