I saw the question up. In fact this has a positive answer and it is a bit of a nice one. 
As mentioned and well known you basicaly  have $Ay=x$ with $y= \begin{pmatrix}0 \\ 1 \\ \vdots \\ n-1\end{pmatrix}$ and $x = \begin{pmatrix} x_1\\ x_2 \\ \vdots \\ x_n\end{pmatrix}$ with $A$ doubly stochastic. This implies that $x\prec y$. For an easier  presentation we rearrange everything so that $x_1\ge\cdots\ge x_n$ and $y_1\ge\cdots \ge y_n$; you want $A$ to verify $Ay=x$ and the condition:
$$(\star) \quad\sum_{j=1}^ka_{i,j}\ge \sum_{j=1}^ka_{i+1,j},\quad \mbox{for all } 1\le i\le n-1 \mbox{ and } 1\le k\le n$$ Here we already take $x$ and $y$ random non negative vectors with $x\prec y$ $(Ay=x)$. 
A sketch of proof is as follows:
1) First if $A$ and $B$ are two doubly stochastic matrices verifying $(\star)$ then $A\cdot B$ also verifies $(\star)$.
2) By known results on majorisation if $x\prec y$ then there is a sequence of vectors $z_i$ with $x=z_0\prec z_1\prec \cdots\prec z_i\prec z_{i+1}=y$ and such that two consecutive vectors differs by only two entries. 
3) From these two it is sufficient to prove that such $A$ exist when $x$ and $y$ differs by only two entries. 
Up to a direct sum of identity matrices we take  $y=\begin{pmatrix}r+h\\r\\\vdots\\r\\r-k\end{pmatrix}$ and $x=\begin{pmatrix}r+h-s\\r\\\vdots\\r\\r-k+s\end{pmatrix}$ and $h\ge s>0$, $k\ge s>0$, with $m$ common terms as $r$,$m\ge 1$ the case $m=0$ is easy.
One can set $$A=\begin{cases}a_{1,1}=a\\a_{i,1}=c, 2\le i\le m+1\\a_{m+2,1}=1-a-mc\\a_{1,j}=u, 2\le j\le m+1\\a_{1,m+2}=1-a-mu\\a_{i,j}=v, 2\le i\le m+1 \,\& \,2\le j \le m+1\\a_{m+2,j}=1-u-mv, 1\le j\le m+1\\a_{i,m+2}=1-c-mv, 2\le i\le m+1\\a_{m+2,m+2}=1-m(1-c-mv)-(1-a-mu)\end{cases}$$
Solving $Ay=x$ in terms of $h,k, u,v$ and $s$ gives the expressions of $a$ and $c$. The condition $(\star)$ and the non negativity for the entries of $A$ reduce to a system of seven (linear) inequalities, in the plane $(u,v)$, with a non empty set of solutions.