$R>r>0$ , $||a||,||b||R$ ; pulls back volume form to itself? Let $n>1$ , consider $\mathbb R^n$ with standard volume form $\omega _0 =dx_1 \wedge ...\wedge dx_n$ ; let $r,R$ be two positive real numbers with $R>r$ . Then is it true that for every $a,b \in B(0;r)$ , $\exists$ a diffeomorphism $f:\mathbb R^n \to \mathbb R^n $ such that $f(a)=b ; f(x)=x , \forall x \in \mathbb R^n $ with $||x||>R$ and $f^*\omega _0=\omega _0$ ? 
As $f^*(dx_1 \wedge ...\wedge dx_n)=(det f' )(dx_1 \wedge ...\wedge dx_n)$ , so basically given $R>r>0$ and $||a||,||b||<r$ , we have to find a diffeomorphism $f : \mathbb R^n \to \mathbb R^n  $ with $f(a)=b$ ; $f|_{\mathbb R^n \setminus B[0;R]}=Id$ and $det f'(p)=1 , \forall p \in \mathbb R^n $ .
Definitely this cannot work out in $\mathbb R$ , because the only differentiable function satisfying $det f'(p)=1 , \forall p$ and $f(a)=b$ is $f(x)=x+b-a$ and this function never gives identity unless $a =b$
 A: I'm going describe my "codifferential" idea in terms of vector fields using coordinates $(x,y_2,\ldots,y_n).$ Let $r^2 = x^2 + |y|^2$ be the usual radial coordinate and let $$\psi(r) = \begin{cases}\exp\left(-\frac{1}{1-r^2}\right) & r<1 \\ 0 & \text{otherwise} \end{cases}.$$
Define the vector field $$V = \sum_{k=2}^n \left( (\psi + y_k \frac{\partial \psi}{\partial y_k}) e_x - \frac{\partial \psi}{\partial x}y_k e_{y_k} \right).$$
(I found this by taking the codifferential of the $2$-form $\sum_k \psi y_k dx \wedge dy_k$.)
Here's a slice of this $V$ for $n=3$, which hopefully communicates the whole picture since it's rotationally symmetric about the $x$ axis:

You should be able to confirm the following facts about $V$:


*

*$V = 0$ outside the unit ball and $V$ is smooth everywhere;

*$V$ is divergence-free;

*When restricted to the $x$-axis, $V$ is a non-vanishing vector field tangent to this axis.


Point 1 implies that the flow of $V$ generates a family of diffeomorphisms $f_t$ that are the identity outside the unit ball. Point 2 implies that each $f_t$ is volume-preserving. Point 3 implies that for any point $p \in B_1$ that lies on the $x$-axis, there is a $t$ such that $f_t(0) = p$.
Conjugating this with rotations (which are volume-preserving diffeomorphisms) lets you send $0$ to any point in the unit ball; and conjugating with scaling lets you change the radius of the ball.
A: Consider the problem of switching two antipodal points $p,-p\in\partial B_s$, for an arbitrary radius $s>0$. Take a 1-dimensional circle $S$ centered at $0$ passing through $p,-p$
and take
$$U:=\left\{q\in\mathbb{R}^n:\text{dist}(q,S)<\frac{s}{2}\right\}. $$
You can find a map $\chi:S^1\times B_{s/2}^{n-1}\to U$ in such a way that $S=\chi(S^1\times\{0\})$ and $\chi^*\omega_0=h(y)\,dt\wedge dy_1\wedge\dots\wedge dy_{n-1}$ ($S^1\times B_{s/2}^{n-1}$ being parametrized by the variables $t\in\mathbb{R}/2\pi\mathbb{Z},y_1,\dots,y_{n-1}$): just identify the slice around $p$, namely $\{q\in U:|q-p|=\text{dist}(q,S)\}$, with $\{0\}\times B_{s/2}^{n-1}$ and for any $p'\in S$ use the unique orientation-preserving rotation which carries $p\mapsto p'$ and equals the identity on $S^\perp$.
Now take any cut-off function $\phi\in C^\infty_c(B_{s/2}^{n-1})$ with $\phi(0)=1$. The map
$$ F:S^1\times B_{s/2}^{n-1}\to S^1\times B_{s/2}^{n-1},\quad F(t,y):=(t+\pi\phi(y),y) $$
preserves $\chi^*\omega_0$, so $f:=\chi\circ F\circ\chi^{-1}$ preserves the volume and switches $p$ with $-p$.
In order to switch your points $a$ and $b$, take a chain $a_0,\dots,a_N$ where $a_i:=a+\frac{i}{N}(b-a)$ and $N>\frac{3|b-a|}{4(R-r)}$.
You can switch two adjacent points in this chain as above, using $s:=\frac{|b-a|}{2N}$ and conjugating with a translation, by means of a diffeomorphism which equals the identity outside $B_{r+(3/2)s}\subseteq B_R$. So you can switch $a$ and $b$ by composing $2N-1$ such diffeomorphisms.
