Stochastic matrix A^n=const. Let $A\in M(n,n)$ be a doubly stochastic matrix such that
$$
A^k=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n\\
\end{pmatrix}
$$
for some $k>1$. Does it follow that
$$
A=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n\\
\end{pmatrix}?
$$

No, it does not follow that
$$
A=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n
\end{pmatrix}.
$$
Here is a (minimal) counterexample. Take
$$
J_3=\begin{pmatrix}
1/3 & 1/3 & 1/3\\
1/3 & 1/3 & 1/3\\
1/3 & 1/3 & 1/3
\end{pmatrix},\qquad N=\begin{pmatrix}
1 & 1 & -2\\
-1 & -1 & 2\\
0 & 0 & 0
\end{pmatrix}.
$$
Observe that $N^2=J_3N=NJ_3=0$. Consequently for all $\varepsilon$,
$$
(J_3+\varepsilon N)^2=J_3^2=J_3.
$$
When $\varepsilon\leq 1/6$, $J_3+\varepsilon N$ is doubly stochastic, producing a counterexample in the case $n=3,k=2$.
