A particle moves in a circle from points $0,1,2,3,4$ (in a clockwise order) it moves one step clock wise with probability $q$ and one step counter clockwise with probability $(1-q)$.
Let $X_n$ denote its location in the circle ($\{X_n:n\geqslant 0\}$ is a Markov chain).
How do I find the steady state probabilities considering the particle is always moving?
I started by finding the transition matrix but I don't know where to go from there;
$$\begin{pmatrix}0&q&0&0&1-q\\ 1-q&0&q&0&0\\ 0&1-q&0&q&0\\ 0&0&1-q&0&q\\ q&0&0&1-q&0\end{pmatrix}.$$
Because this is a doubly stochastic matrix (both the rows and the columns sum to one, the unique stationary distribution is the uniform distribution over $\{0,1,2,3,4\}$. Indeed, \begin{align} &\quad\frac15\begin{pmatrix}1&1&1&1&1\end{pmatrix}\begin{pmatrix}0&q&0&0&1-q\\ 1-q&0&q&0&0\\ 0&1-q&0&q&0\\ 0&0&1-q&0&q\\ q&0&0&1-q&0\end{pmatrix}\\ &=\begin{pmatrix}0&q&0&0&1-q\\ 1-q&0&q&0&0\\ 0&1-q&0&q&0\\ 0&0&1-q&0&q\\ q&0&0&1-q&0\end{pmatrix}. \end{align}