Suppose the state space is $\mathbb N\cup\{0\}$ with transition probabilities
\begin{align}
\mathbb P(X_{n+1}=i+1\mid X_n = i) =: p_i,\quad i=0,1,\ldots\\
\mathbb P(X_{n+1}=i\mid X_n = i) =: r_i,\quad i=0,1,\ldots\\
\mathbb P(X_{n+1}=i-1\mid X_n = i) =: q_i,\quad i=1,2,\ldots\\
\end{align}
For this to give a valid transition matrix $P$, we must have $r_o+p_0=1$ and $q_i+r_i+p_i=1$ for $i\geqslant 1$. If $\nu(0)=1$ and $\nu(n) = \prod_{i=1}^n\frac{p_{i-1}}{q_i}$ then $\nu$ is an invariant measure for $P$ iff $\sum_{n=0}^\infty \nu(n):=C<\infty$. In this case, there is a unique stationary distribution $\pi$ given by $\pi = \frac1C\nu$.
For $P$ to be a doubly stochastic matrix, we must have $r_o+q_1=1$ and $p_{i-1}+r_i+q_{i+1}=1$ for $i\geqslant 1$. Now, to have $\sum_{n=0}^\infty \nu(n)<\infty$, we must have
$$
\liminf_{n\to\infty} \frac{p_n}{q_{n+1}}<1
$$
(consider the case where the $p_n$ and $q_n$ are constant and so the sum is a geometric series). However, from $r_0+p_0=1$ and $r_0+q_1=1$ we get $p_0=q_1$, and from $q_i+r_i+p_i=1$ and $p_{i-1}+r_i+q_{i+1}=1$ we get $p_{i-1}+q_{i+1} = p_i+q_i$. Now, by induction we see that $p_{i-1}=q_i$, and hence $p_i=q_{i+1}$, for all $i\geqslant 1$. This implies that $p_{i-1}=q_i=\frac12$ for all $i$, and so if $P$ is doubly stochastic, it necessarily cannot have a stationary distribution (for it is null recurrent).
