Let $A$ be an $n \times n$ nondegenerate square matrix with real-valued entries. If we interpret the rows of $A$ as points in $\mathbb{R}^n$, then $A$ defines a simplex. We'll say $v \in \mathbb{R}^n$ is a normal vector to $A$ if $v$ is the normal vector to the hyperplane on which this simplex lies.
I am looking for a function $f_A: \mathbb{R}^n \to \mathbb{R}^{n \times n}$ that maps a normal vector to a rotation of $A$ that has that normal vector. It's easy to see that many rotations of $A$ might correspond to a single normal vector; thus, many implementations of $f_A$ might be possible. Any of them will do, as long as $f_A$ is continuous.
Thanks!
If I've understood the question correctly, this is impossible for odd $n$.
Any rotation of $A$ transforms its unit normal vector $v$ by the same rotation, so what you're really looking for is a continuous function that, for any vector $u$, gives a rotation matrix $R(u)$ which maps $v$ to $u$.
Let $w$ be a fixed unit vector orthogonal to $v$. Then $R(u)w$ is orthogonal to $R(u)v = u$. Now $R(u)w$, seen as a function over the unit sphere, defines a vector field which is everywhere tangential to the sphere and nowhere zero. By the hairy ball theorem, such a vector field cannot be continuous when $n$ is odd, so $R(u)$ cannot be either.