Find linear map $A$ with $\det A = 1$ such that $y = Ax$ Is it possible to explicitly construct a linear map (matrix) $A \in \mathbb R^{n \times n}$ such that $\det A = 1$ and $y = Ax$ for the given vectors $x,y \in \mathbb R^n$
In $\mathbb R^3$ you can find the (unique?) rotation axis $a \times b$ (cross product) and you can find the angle by using the scalar product, and I am not sure if we need $|x|=|y|$ (I thought perhaps (figuratively) you can 'squish' one dimension and 'stretch' another in order to still get $\det = 1$?
 A: Yes, if both $x$ and $y$ are nonzero, and $n>1$. If both $x$ and $y$ are zero, of course any special linear transformation works. If one is zero and one is nonzero, then no nonsingular transformation will map one to the other, much less a special linear transformation.
First suppose $y=\lambda x$ and that $x$ is the first element of the basis you are using. 
Then $\begin{bmatrix}\lambda&0&0\\
0&\lambda^{-1}&0\\
0&0&I_{n-2}\end{bmatrix}$ is a special linear matrix sending $x$ to $y$.
If $x$ and $y$ are linearly independent, use them as the first two elements of your basis in that order. 
Then $\begin{bmatrix}0&1&0\\
-1&0&0\\
0&0&I_{n-2}\end{bmatrix}$ is a special linear matrix sending $x$ to $y$.
A: For the case where $x$ and $y$ are linearly independent, complete them to a basis $\{x,y,z_3,z_4,\dots,z_n\}$. Then consider a transformation such that $x\mapsto y$, $y\mapsto x$ and $z_3\mapsto -z_3$ and $z_k\mapsto z_k$ for $k\ge 4$; its matrix with respect to the given basis is
$$
\begin{bmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & I_{n-3}
\end{bmatrix}
$$
Of course, if $n=3$ the lower right angle is not to be considered.
The case when $y=\lambda x$ has already been covered, but just for completeness, consider a basis $\{x,z_2,z_3,\dots,z_n\}$. The map with $x\mapsto \lambda x=y$, $z_2\mapsto \lambda^{-1}z_2$ and $z_k\mapsto z_k$ $(k\ge2)$ will do.
Actually, this case works also for $n=2$. In case $n=2$ and $\{x,y\}$ is a basis, you must have $x\mapsto y$ and $y\mapsto \alpha x+\beta y$, so the matrix is
$$
\begin{bmatrix}
0 & \alpha\\
1 & \beta
\end{bmatrix}
$$
and taking $\alpha=-1$ and $\beta=0$ will do.
