Let $G$ be a finite group. Let $A<G$.
When is it true that $G$ has only a single maximal subgroup containing (or equal to) $A$?
The necessary condition is that there must be $x \in G$ such that $\left<A \cup \{x\}\right>=G$ and the order of $\left<x\right>$ is a power of a prime. In many cases, this is also sufficient -- for instance, if $G$ is abelian. But this is not generally true. The simplest counterexample is provided by $G=S_2 \times C_3$, $A=\left<(e,v)\right>$ where $e$ is the identity and $v$ is any $2-$cycle permutation.
Unless I'm missing something, it is sufficient for there to be $x \in G$ such that $\left<A \cup \{x\}\right>=G$, the order of $\left<x\right>$ is a power of a prime, and either $\left<x\right>$ or $A$ are normal subgroups of $G$. But this condition is not necessary if $G$ is simple nonabelian and $A$ is any of its maximal subgroups, for instance.
EDIT: a slightly weakened version of the sufficient condition presented above is that there is $x \in G$ such that $\left<A \cup \{x\}\right>=G$, the order of $\left<x\right>$ is a power of a prime, and $A$ commutes with $\left<x\right>$ under the product of groups. However, this is still not necessary. In particular, there are examples where the supergroups of $A$ do not form a chain, for instance if $A≅C_2$ is a non-normal subgroup of $G=C_2^2 ⋊C_4$.
