$\newcommand{\R}{\mathbb{R}}$Suppose $\Omega$ is a bounded, convex domain in $\R^{m}$. Fix $x_1, x_2\in\Omega$ and an invertible matrix $A\in\mathrm{GL}^{+}(m)$ with positive determinant. Let $U\subset\Omega$ be an open convex neighbourhood containing $x_1, x_2$ and the Euclidean geodesic connecting them.

I want to construct a smooth diffeomorphism $\tau\in\mathrm{C}^{\infty}(\Omega;\Omega)$ of $\Omega$ such that $$ \begin{align} \tau(x_1)&=x_2 \tag{a}\\ \mathrm{d}\tau_{x_1} &= A \tag{b}\\ \tau|_{\Omega\setminus U} &\equiv \operatorname{Id} \tag{c} \end{align} $$

This seems (to me at least) like an underdetermined problem which should have a solution. How can one construct $\tau$? Am I completely mistaken and the boundedness of $\Omega$ is a topological obstruction to the existence of such a diffeomorphism?

I've tried the following:

- Let $r>0$ be so small that $\overline{B_r(x_1)}\subset U$ and consider a smooth bump function for $\overline{B_r(x_1)}$ supported in $U$, i.e. $\chi$ such that $0\leq\chi\leq1$, $\chi\equiv1$ on $\overline{B_r(x_1)}$ and $\chi\equiv0$ on $\Omega\setminus U$. Consider the affine transformation $T$, defined by $T(x)=x_2 + A(x-x_1)$. If need be let's make $r$ so small that the ellipsoid $x_2 + A(B_r(x_1))$ is contained in $U$ as well. Define $\tau$ as the convex combination $\chi \cdot T + (1-\chi) \cdot \operatorname{Id}$. This is a smooth map and fulfils $(a)-(c)$ but I believe it is not a diffeomorphism -- even in the case where all eigenvalues of $A$ are positive.
- Unsuccessfully tried finding a vector field by constructing flow lines between the points $x_1 + e_1, \ldots x_1 + e_m$ and $x_2 + Ae_1, \ldots x_2 + Ae_m$. I hoped that the (smoothly cutoff) flow would be the sought after diffeomorphism.
- Unsuccessfully tried finding an energy functional whose gradient flow induces the diffeomorphism.
- Might it be possible to write down an equation that is solved by applying the implicit function theorem (for Banach spaces, so $\tau\in\mathrm{C}^{k}(\Omega;\Omega)$)?
- I think I can reduce it to the case $x_1=x_2$ if that simplifies things.