The diffeomorphism of $\mathbb R^n$

If $f$ is a diffeomorphism of $\mathbb R^n$ and $K$ is a compact set in $\mathbb R^n$, can we find another diffeomorphism $\tilde f$ of $\mathbb R^n$ such that:

(1)$f=\tilde f$ on a neighborhood of $K$. (2)There is a bounded set $V$ and $\tilde f=id$ outside $V$?

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The function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=-x$ and $K=[-1,1]$ provides a counterexample - any continuous map $\tilde{f}$ satisfying properties 1) and 2) would necessarily fail to be injective by the intermediate value theorem.
As Jason DeVito points out below, we can use a similar setup to create a counterexample for any $n$. Letting $f:\mathbb{R}^n\to\mathbb{R}^n$ be defined by $f(x_1,x_2,\ldots,x_n)=(-x_1,x_2,\ldots,x_n)$, we have for all $p\in\mathbb{R}^n$ $$\det(df_p)=\begin{vmatrix} -1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{vmatrix}=-1.$$ Thus, any diffeomorphism $\tilde{f}$ satisfying 1) would also have to have $\det(d\tilde{f}_p)=-1$ for some $p\in\mathbb{R}^n$. Because $\tilde{f}$ is a diffeomorphism, it can't have $\det(d\tilde{f}_p)=0$ for any $p\in\mathbb{R}^n$. Because $\det(d\tilde{f}_p)$ varies continuously with $p$, we must have $\det(d\tilde{f}_p)<0$ for all $p\in\mathbb{R}^n$. But $\det(dI_p)=1>0$ for all $p\in\mathbb{R}^n$, so we can't have $\tilde{f}=I$ anywhere.
Your answer can be promoted to $n$ by the same kind of reasoning. Pick your favorite orientation reversing diffeomorphism $f$ and let $K$ be a cube. Then $f:K\rightarrow f(K)\subseteq \mathbb{R}^n$ reverses orientation, so the determinant of the differential is negative. Since the determinant is always nonsingular, it must, thus, be negative everywhere. Hence, $\tilde{f}$ can't be the identity outside $V$. – Jason DeVito Apr 28 '12 at 3:41
+1 Excellent point! You are certainly welcome to post that as a separate answer, or I can incorporate it into my post. Hmm, now I wonder if the statement becomes true if we require $f$ to be orientation-preserving, then. – Zev Chonoles Apr 28 '12 at 3:45
A modified form of the function you gave should show that it doesn't work in $\mathbb{R}^n$, either. I think $f(x',x_n) = (x', -x_n)$, on the box $[-1,1]^n$ is a counterexample. Maybe we could also pick $f(x) = -x$ on $[-1,1]^n$, but I think this requires that $n$ is odd. – Nicholas Stull Apr 28 '12 at 3:49
If $f$ is orientation-preserving and there is a $\epsilon>0$ such that $|f-id|<\epsilon$, then the conclusion holds. – Hezudao Apr 28 '12 at 12:03