TITLE: Can a diffeomorphism between submanifolds be extended to a self-diffeomorphism of total manifold?
QUESTION [8 upvotes]: Let $M^{n}$ be a differential manifold and $S$, $N$ submanifolds of $M$ with same dimension $\lt n$, $\phi$ is a diffeomorphism from $S$ to $N$, can $\phi$ be extended to a self-diffeomorphism of $M$? Or does there exist such $\phi$, $\phi$ can be extended?
Special case is $M=\mathbb{R}^{n}$, $S$ is an $s$-dimensional submanifold $(s\le n)$,and $\phi$ is a diffeomorphism from $S$ to an open set in $R^{s}$, the question is can $\phi$ be extended to a self-diffeomorphism of $\mathbb{R}^{n}$? If not, does there exist such $\phi$, $\phi$ can be extended?

REPLY [14 votes]: Let $M = \mathbb{R}$ and $S = N = \{0, 1, 2\}$. 
Consider the diffeomorphism $\varphi : \{0, 1, 2\} \to \{0, 1, 2\}$, given by $\varphi (0) = 0$, $\varphi(1) = 2$, $\varphi(2) = 1$. As every diffeomorphism $\mathbb{R} \to \mathbb{R}$ is monotonic, there is no diffeomorphism $\psi : \mathbb{R} \to \mathbb{R}$ such that $\psi|_{\{0, 1, 2\}} = \varphi$.