The distribution isn't exactly binomial.. I'll explain what I mean: I am considering a Markov chain with N+1 states (denoted by $0,1,...,N$), the probability of going from state $i$ to state $j$ is given by
$$Pr[i,j]={N \choose j}\left(\frac i N\right)^j\left(\frac{N-i}N\right)^{N-j}.$$
Furthermore, since by this assignment the first and last rows would consist of $0'$s, we assign
$$Pr[0,j]=\frac1{N+1}$$
and
$$Pr[N,j]=\frac1{N+1}.$$
That is, from states $0$ or $N$, the probability of going to any other state is uniform.
What I am looking for is a row vector which when multiplied on the right by the transition matrix defined above, will yield the same vector.
Clearly there are certain properties that a stationary distribution will have: Since
$$Pr[i,i]=Pr[(N-i),(N-i)],$$
we know that the matrix, and therefore the stationary distribution must be symmetric. For this reason I have assumed that the first and last entries in the vector are 1 (since the other entries can always be scaled appropriately) I have tried the standard method of computing the eigenvectors corresponding to the eigenvalue $1$, and have managed to come up with stationary distributions when $N=3$ and $4$, given by $\left(1, \frac32, \frac32, 1\right)$ and $\left(1, \frac{224}{145}, \frac{244}{145}, \frac{224}{145}, 1\right)$ resp.
I am really not sure how to proceed with this problem and would really appreciate if anyone could give me some insight into this type of problem. Let me know if there are any additional details I could provide. Thanks!