$L^2$ Bounds for Markov Chains. Consider a non-negative, square stochastic $n \times n$ matrix $P$ (rows sum to one, $P$ is ergodic).
We are interested in characterizing the set of $n \times n$ invertible matrices $A$ such that we have:
$\| APA^{-1} \|_2 \leq 1$
Here $ \| \cdot  \|_2$ means a matrix norm induced by the vector norm $L_2$.
One example is $\operatorname{diag}(\xi)^{1/2}$, where $\xi$ is the vector representing the stationary distribution and $\operatorname{diag}$ is an operator constructing a diagonal matrix with zeros off the diagonal.
I am interested in a rule defining all (or at least a large class) of valid choices for $A$.
 A: This is not a full answer, but maybe helpful anyway, as it clarifies, that an answer depends on the properties of $P$ itself.
The induced matrix norm of the $L_2$-vector norm is the spectral norm, which is the maximal singular value of the matrix under consideration, so in order to find out something about $||APA^{-1}||_2$, we should look at
$$(A^{-1})^TP^TA^TAPA^{-1} \, .$$
If $A$ is orthogonal, so $A^{-1} \, = \, A^T$, this expression simplifies to
$$ AP^TPA^{-1} \, ,$$
while if $A$ commutes with $P$, so $AP \, = \, PA$, it simplifies to 
$$ P^TP \, .$$
Therefore, as $P^TP$ and $AP^TPA^{-1}$ have the same spectrum, we have to find out something about the Perron-eigenvalue of $P^TP$, given $P$ is row stochastic.
By now, the nice answer of user1551 (my own answer was really clumsy compared to that) to your question
Characterize stochastic matrices such that max singular value is less or equal one. 
gives
$$||APA^{-1}||_2 \ \geq \ 1$$
for row stochastic $P$ and orthogonal or commuting $A$ and 
$$||APA^{-1}||_2 \ = \ 1 \, ,$$
if and only if $P$ is doubly stochastic. Still, I leave my example here:
If $P$ is row stochastic, $P^T$ is column stochastic, which implies, that $P$ and $P^TP$ have the same column sums, because any matrix, which is column stochastic, preserves the component sum of any vector multiplied with it from the right.
Therefore, if $P$ is doubly stochastic, $P^TP$ is also doubly stochastic and the orthogonal matrices or matrices, which commute with $P$ are included in the class of matrices you are looking for in this case.
If $P$ is not doubly stochastic, there is at least one column sum larger than $1$, while the column sums sum to $n$. This always implies the Perron-eigenvalue of $P^TP$ to be larger than $1$, see the answer of user1551 mentioned above, example:
\begin{equation}
\mathbf{P}  \ := \ 
\begin{pmatrix} 
\frac{1}{3}  &  \frac{1}{3}   & \frac{1}{3}  \\
    1        &         0      &      0       \\
    0        &         0      &      1       
\end{pmatrix} 
\end{equation}
\begin{equation}
\mathbf{P^TP}  \ := \ 
\begin{pmatrix} 
\frac{10}{9} &  \frac{1}{9}  &   \frac{1}{9}  \\
\frac{1}{9}  &  \frac{1}{9}  &   \frac{1}{9}  \\
\frac{1}{9}  &  \frac{1}{9}  &   \frac{10}{9}  
\end{pmatrix} 
\end{equation}
The Perron-eigenvalue of $P^TP$ is $ \frac{\sqrt{3}+2}{3} \, > \, 1$, so no matrix commuting with $A$ or any orthogonal matrix is contained in the sought class for the matrix $P$.
A: @ ziutek , a stochastic matrix does not necessarily admit a stationary distribution. cf. below.
Note that $||APA^{-1}||_2=\sqrt{\rho (MM^T)}=\max_i \sigma_i$ where $M=APA^{-1}$ and $(\sigma_i)_i$ are the singular values of $M$ in decreasing order. Moreover $\rho(M)=\rho(P)=1$ and (it is true for every matrix) $\sigma_1\geq \rho(M)$. Thus your question is "find $A$ s.t. $||M||_2=\rho(M)$".
In this file $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$ , the following result is shown:
Prop. (user1551): there is a sub-mult. norm $||.||$ s.t. $ρ(U)=||U||$ iff the eigenvalues of $U$ of maximal modulus are semi-simple. 
Necessarily, the eigenvalues of $P$ of  modulus $1$ must be semi-simple. That is the case when $P$ is irreducible (because they are simple) that we'll assume in the sequel (that does not imply that a stationary distribution exists !). We may assume that $P=diag(T,N)$ where $\rho(T)<1$ and $N^h=I$ where $h$ is the period of $P$. Take $A=diag(B,C)$ with $B$ as in the proof above cited; then $||APA^{-1}||_2=||CNC^{-1}||_2$ and the condition is $||CNC^{-1}||_2=1$. I think that we MUST choose $C$ as follows: $C=RS$, a REAL matrix, s.t. $S$ (a complex matrix) diagonalizes $N$ and $R$ is unitary. 
EDIT: A mistake. What I wanted to say is that a stationary distribution is not necessarily a limit distribution.
