Can a diffeomorphism between submanifolds be extended to a self-diffeomorphism of total manifold? 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?
 A: I know that this is a trivial counterexample with a specific $\phi$, but if you're not putting any constraints and let $s\leq n$ then you can also consider two disjoint open bounded intervals $I_1, I_2\subset\mathbb R$ and the diffeomorphism which is the identity on $I_1$ and flips $I_2$. This does not extend because for example the two "inner" extremes must be sent to the same connected component of $\mathbb R\backslash(I_1\cup I_2)$, which is impossible.
A: Consider $S = $ a union of two concentric circles, $N = $ two circles, side by side (e.g., unit circles centered at $(\pm 2, 0)$), and $M = \mathbb R^2$. The complement of $S$ contains two components whose fundamental groups are isomorphic to the integers (the annulus between the two circles, and the unbounded piece); the complement of $N$ contains two components (the disks bounded by the circles) whose fundamental groups are trivial. Since a self-homeomorphism of $M$ that carries $S$ to $N$ would have to carry the complements to the complements, there can be no such self-homeomorphism. 
This is probably as simple an example as you're going to find, for in the line, no such example works. 
A: Consider a standard circle and a trefoil knot in $\mathbb{R}^3$. Any diffeomorphism of $\mathbb R^3$ to itself that took one to the other would have to be a diffeomorphism of the complements. But the fundamental groups of the complements are different, so such a diffeomorphism does not exist. 
A: 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 $\psi|_{\{0, 1, 2\}} = \varphi$.
