What matrices preserve the $L_1$ norm for positive, unit norm vectors? It's easy to show that orthogonal/unitary matrices preserve the $L_2$ norm of a vector, but if I want a transformation that preserves the $L_1$ norm, what can I deduce about the matrices that do this?  I feel like it should be something like the columns sum to 1, but I can't manage to prove it.
EDIT:
To be more explicit, I'm looking at stochastic transition matrices that act on vectors that represent probability distributions, i.e. vectors whose elements are positive and sum to 1.  For instance, the matrix
$$
M = \left(\begin{array}{ccc}1 & 1/4 & 0 \\0 & 1/2 & 0 \\0 & 1/4 & 1\end{array}\right)
$$
acting on
$$
x=\left(\begin{array}{c}0 \\1 \\0\end{array}\right)
$$
gives
$$
M \cdot x = \left(\begin{array}{c}1/4 \\1/2 \\1/4\end{array}\right)\:,
$$
a vector whose elements also sum to 1.
So I suppose the set of vectors whose isometries I care about is more restricted than the completely general case, which is why I was confused about people saying that permutation matrices were what I was after.
Sooo... given the vectors are positive and have entries that sum to 1, can we say anything more exact about the matrices that preserve this property?
 A: The matrices that preserve the set $P$ of probability vectors are those whose columns are members of $P$.  This is obvious since if $x \in P$, $M x$ is a convex combination of the columns of $M$ with coefficients given by the entries of $x$.  Each column of $M$ must be in $P$ (take $x$ to be a vector with a single $1$ and all else $0$), and $P$ is a convex set.
A: Since you originally asked about $L^1$ spaces I dared to add this comment.  
If one wants to preserve the integral in (finite-dimensional and with finite measure ) $L^1$ spaces rather than the norm of $\ell^p$, the matrices $M$ that do this are more general than the stochastic matrices. 
One can define these matrices with two components, labeled $S$ (for the stochastic component) and $G$ (for the generalized permutation matrix component) such that that $M= S * G$, where * represents the Hadamard product. 
The $S$ matrices are effectively stochastic matrices as shown by Robert Israel.  
The $G$ matrix is given by the unique matrix resulting of the outer product $u_{\mu} \otimes \frac{1}{u_{\mu}} := | u_{\mu} \rangle \langle \frac{1}{u_{\mu}} |$ of the unique column vector 
$u_{\mu} :=\left(\begin{array}{c}\mu_1 \\ \mu_2 \\ \ldots \\ \mu_2\end{array}\right)$ and the also unique row vector $\frac{1}{u_{\mu}} :=\left(\frac{1}{\mu_1} \  \frac{1}{\mu_2} \  \ldots \  \frac{1}{\mu_n}\right)$:
$G:=u_{\mu} \otimes \frac{1}{u_{\mu}} = \left(\begin{array}{cccc}
1 & \frac{\mu_2}{\mu_1} & \ldots & \frac{\mu_n}{\mu_1} \\ 
\frac{\mu_1}{\mu_2} & 1 & \ldots & \frac{\mu_n}{\mu_2} \\ 
\ldots & \ldots & \ldots & \ldots \\ 
\frac{\mu_1}{\mu_n} & \frac{\mu_2}{\mu_n} & \ldots & 1 \end{array}\right)$
where $\mu_i$ are the measures of the generating family of subsets $\{ A_i \}$ of the underlying sigma algebra, i.e. $\mu_i := \mu(A_i)$ and $n = |\{ A_i \}|$. 
To give you an example of where the stochastic component $S$ is absent, take the stochastic matrix $S$ to be simply a permutation matrix. In this case your $M$ that preserves the integral is a generalized permutation matrix whose non-zero elements are of the form $A_{i,j} =\frac{\mu_{j}}{\mu_i}$. 
To see why the measure values $\mu_i$ are needed in the definition of $M$ recall that a $L^p$ space is defined given a measure space $(X,\Sigma,\mu)$. So if the $L^1$ space is finite dimensional then the vectors $v$ in $L^1$ are simple functions, whose integral is defined as the product $\langle u_{\mu}|v\rangle$. And if the measure is finite, then this integral is always well defined.
I hope I made myself clear.
