# Given an invariant distribution is the (finite state) Markov transition matrix unique?

Doeblin's theorem states that for a given transition probability matrix there exists a unique invariant distribution for that chain.

Is the converse true as well? Can two (finite state, discrete) Markov Chains have the same invariant distribution, but have different transition matrices?

I don't think so, but I have only tried a proof by contradiction:

Assume there are two transition matrices, P and Q such that $\pi$ P = $\pi$ and $\pi$Q = $\pi$ ($\pi$ is the invariant distribution). Then $\pi$ P = $\pi$Q and P=Q (contradiction).

However, this doesn't really tell me why this should be true.

Unfortunately your proof contains an error, $\pi P = \pi Q$ does not imply $P=Q$. Consider for example $$P = \begin{pmatrix} 0.5 & 0.5 & 0 \\ 0.5 & 0 & 0.5 \\ 0 & 0.5 & 0.5 \end{pmatrix}$$ and $$Q = \begin{pmatrix} 0.33 & 0.33 & 0.33 \\ 0.33 & 0.33 & 0.33 \\ 0.33 & 0.33 & 0.33 \end{pmatrix}.$$ These both have the stationary distribution $\begin{pmatrix} 0.33 & 0.33 & 0.33 \end{pmatrix}$, but are two different matrices.