For any $n \in \mathbb{N}$, the collection $\mathcal{P}_n$ of permutation matrices is obtained by permuting the rows of the $n \times n$ identity matrix. Under the operation of matrix multiplication, $\mathcal{P_n}$ is a subgroup of $\mathcal{G_n}$ -- the subgroup of $n \times n$ orthogonal matrices $\mathcal{O}_n$ that keeps the vector $e = (1, \ldots, 1)^t \in \mathbb{R}^n$ fixed.
Note that $\mathcal{G_n} \cong \mathcal{O}_{n-1}$. Is $\mathcal{G_n}$ the smallest infinite Lie subgroup of $\mathcal{O}_n$ containing $\mathcal{P_n}$?
Basically, can $\mathcal{G_n}$ be a "continuous" extension of $\mathcal{P_n}$?
Let $G$ be an infinite Lie group satisfying $\mathcal{P}_n\le G\le\mathcal{O}_{n-1}$. Since $\mathcal{O}_{n-1}$ is compact, any $0$-dimensional subgroup is finite. Therefore, $G$ has dimension $\ge 1$. This implies that the Lie algebra $\mathfrak{g}=\text{Lie}(G)$ contains a nonzero element $\gamma\in\mathfrak{g}\subseteq\mathfrak{o}_{n-1}$. The set $\{\text{Ad}_P(\gamma):P\in\mathcal{P}_n\}$ spans $\mathfrak{o}_{n-1}$, so $\mathfrak{o}_{n-1}\subseteq\mathfrak{g}$ and thus $\mathfrak{g}=\mathfrak{o}_{n-1}$. Since $\mathcal{P}_n$ (and thus $G$) contains an element in each connected component of $\mathcal{O}_{n-1}$, this implies $G=\mathcal{O}_{n-1}$. Thus there are no infinite proper Lie subgroups of $\mathcal{O}_{n-1}$ that contain $\mathcal{P}_n$.
$\mathcal{O}_{n-1}$ is an extension of $\mathcal{P}_n$, and it is in a limited sense "locally optimal", but there are other extensions. It would be natural, for instance, to require that the extension be compact and connected, and while it's still difficult to find a tenable notion of optimality, at least each finite groups has such an extension.