TITLE: What are the matrices preserving the $\ell^1$-norm?
QUESTION [17 upvotes]: So I am inspired by unitary matrices which preserve the $\ell^2$-norm of all vectors, so in particular the unit norm vectors. But then I saw that the $\ell^1$-norm of probability vectors is preserved by matrices whose columns are probability vectors. And this got me thinking: But what are the matrices preserving the $\ell^1$-norm of arbitrary real unit $\ell^1$-norm vectors? So basically we extend a probability vector to also allow a sign, but ignoring the signs, this should still be a probability vector; and then we ask for the corresponding structure-preserving matrices.
It is already clear that the columns of such a matrix should be this 'extended' kind of probability vector, because we can multiply the matrix with a standard basis vector which has $\ell^1$-norm 1. But not all of such matrices preserve this, take for example
$$ M = \frac{1}{2} \left(\begin{matrix} 1 & 1\\ 1 & -1 \end{matrix}\right) $$ 
and 
$$ x = \left( \begin{matrix} 0.3 \\ -0.7 \end{matrix} \right) $$ 
Then we have 
$$ Mx = \left(\begin{matrix} -0.2 \\ 0.5 \end{matrix}\right) $$ 
which fails the test.

REPLY [4 votes]: Here is yet another (sketch of the) proof that the matrices preserving the $p$-norm for $p\neq 2$ are generated by permutation matrices, and the diagonal ones with diagonal elements of absolute value $1$. 
$\let\eps\varepsilon$I address the real case; but the complex one should be similar. 
If $2<p<\infty$, then the tangent hyperplanes to the unit sphere at the points $\pm e_i$ approximate the ball uniformly with the error $\Theta(\eps^p)$. This does not happen at other points; hence these points are permuted, and the matrix is a permutation matrix (with, possibly, changed signs).
Similarly, if $1<p<2$, those tangents approximate the ball in the worst possible way, thus the same result.
For $p=1$ or $p=\infty$, the vertices of the ball map to the vertices, and the edges to the edges; hence the result follows again.