Let $S$ be a naked set (later this will be equipped with various smooth structures, making it into various smooth manifolds, $\mathcal{M}_0$, $\mathcal{M}_1$, $\dots$). Let $H_1\subset\text{Perm}(S)$ be a (potentially infinite-dimensional) Lie group which acts transitively on $S$. Let $H_2\subset H_1$ be the stabilizer subgroup of $H_1$'s action on $S$.
The following result is true if $H_1$ is a finite-dimensional Lie group and I would like to know if this claim also holds in the infinite-dimensional case.
Finite Dimensional Result: Suppose that there exists a smooth structure for $S$ (e.g., a maximal atlas, $\mathcal{A}_0$) such that $\mathcal{M}_0=(S,\mathcal{A}_0)$ is a smooth finite-dimensional manifold with the following property: The action of each $h\in H_1$ is smooth, i.e., $H_1\subset\text{Diff}(\mathcal{M}_0)$. Claim: If such a smooth structure $\mathcal{M}_0$ for $S$ exists, then it is unique up to diffeomorphism.
Proof: Suppose that, in addition to $\mathcal{M}_0=(S,\mathcal{A}_0)$, some other smooth structure $\mathcal{M}_1=(S,\mathcal{A}_1)$ has $H_1\subset\text{Diff}(\mathcal{M}_1)$. Note that $\mathcal{M}_1$ is homogeneous with respect to $H_1$ (i.e., its action is smooth and transitive). Hence, we can reconstruct $\mathcal{M}_1$ up to diffeomorphism as $\mathcal{M}_1\cong H_1/H_2$. But we can do the same reconstruction for the original smooth manifold, $\mathcal{M}_0$. Note that $H_2$ is the stabilizer subgroup of $H_1$ no matter what smooth structure we put on $S$. Hence we have $\mathcal{M}_0\cong H_1/H_2$ as well as $\mathcal{M}_1\cong H_1/H_2$ and therefore $\mathcal{M}_1\cong \mathcal{M}_0$.
Example 1) Suppose that $H_1\cong\text{SO}(3)$ acts transitively on a set $S$ with stabilizer subgroup $H_2\cong\text{SO}(2)$. In this case, there is a unique smooth structure $\mathcal{M}$ for $S$ which makes the action of $H_1$ on $S$ smooth. Namely, we must have $\mathcal{M}\cong \text{SO}(3)/\text{SO}(2) = S^2$, the 2-sphere.
Example 2) Suppose that $H_1$ acts transitively on a set $S$ in such a way that its stabilizer subgroup, $H_2\subset H_1$, is not a closed subgroup of $H_1$. Then there is no smooth structure for $S$ which makes the action of $H_1$ on $S$ smooth. But this is okay because the above claim is that existence implies uniqueness.
As I mentioned above, I'd be interested to know if this result also holds for infinite-dimensional Lie groups. For instance, consider the double torus $\mathcal{M}=\mathbb{T}^2+\mathbb{T}^2$. As Mostow (2005) proves, no finite-dimensional Lie subgroup of $G\subset\text{Diff}(\mathcal{M})$ acts transitively on $\mathcal{M}$. Note however, that the infinite-dimensional Lie group $\text{Diff}(\mathcal{M})$ act transitively on $\mathcal{M}$. It is likely that some infinite-dimensional Lie subgroups $G\subset\text{Diff}(\mathcal{M})$ acts transitively on $\mathcal{M}$ as well.
Example 3?) Suppose that I have in hand an infinite-dimensional Lie group $H_1$ which acts transitively on $S$ with stabilizer subgroup $H_2$. Suppose that I know that I can grant $S$ a smooth structure such that it becomes $\mathcal{M}_0\cong\mathbb{T}^2+\mathbb{T}^2$ and moreover that the action of $H_1$ on $S$ is smooth according to $\mathcal{M}_0$. Could there be some other smooth structure $\mathcal{M}_1$ for $S$ which considers $H_1$ smooth but has $\mathcal{M}_1\not\cong \mathbb{T}^2+\mathbb{T}^2$? If $H_1$ were finite-dimensional this would be impossible. What changes if anything when $H_1$ is an infinite-dimensional Lie group?