Two Primitive Stochastic Matrices with Eventually Equal Sequence of Powers Let $A$ and $B$ be two $n\times n$ primitive row-stochastic matrices. That is, all of their entries are non-negative, all the rows sum up to $1$, and there is an integer $p\geq 1$ such that all the entries of $A^p$ and $B^p$ are strictly positive. Assume that $A_{i,j}=0\iff B_{i,j}=0$ for all $i,j$. Here is my question:

Suppose that there exists $k_0\geq 1$ such that $A^k=B^k$ for all
  $k\geq k_0$. Does this imply that $A=B$?

Remark: Note that if any of $A$ and $B$ is invertible, then the answer is positive. Say for instance that $B$ is invartible, then
$$AB^{k_0}=AA^{k_0}=A^{k_0+1}=B^{k_0+1}=BB^{k_0}.$$
Multiplying by $B^{-k_0}$ we get that $A=B$.

The specified conditions are not sufficient to force $A=B$.

Here's an example . . .

Choose $a,b\in (0,\frac{1}{2})$ with $a\ne b$, and let $A,B$ be given by
$\\[5.5pt]$
$$
A = 
\pmatrix
{
a &\frac{1}{2}-a&\frac{1}{2}\cr
a &\frac{1}{2}-a&\frac{1}{2}\cr
\frac{1}{2}-a&a&\frac{1}{2}
}
\\
$$
$$
B = 
\pmatrix
{
b &\frac{1}{2}-b&\frac{1}{2}\cr
b &\frac{1}{2}-b&\frac{1}{2}\cr
\frac{1}{2}-b&b&\frac{1}{2}
}
$$
Then $A\ne B$, but for all $k\ge 2$, we have
$$
A^k=B^k=
\pmatrix
{
{\large{\frac{1}{4}}}\;&{\large{\frac{1}{4}}}\;&{\large{\frac{1}{2}}}\cr
{\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}\cr
{\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}
}
$$
