We can describe the probabilities for the begin state as the vectors $$\mathbf{x}_0 = [0.5,0,\dots,0,0.5]$$ and $$\mathbf{y}_0 = [0,0.5,\dots,0.5,0]$$

If the proces can be coupled then these probabilities must approach each other.

Case $\exists t : \mathbf{x}_t = \mathbf{y}_t$

The probabilities for the states after some time $t$ are must be equal $\mathbf{x}_t = \mathbf{y}_t$ such that

$$ \mathbf{P}^t \mathbf{x}_0 =\mathbf{P}^t \mathbf{y}_0 .$$

And also

$$\mathbf{P}\cdot \left(\mathbf{P}^{t-1} (\mathbf{x}_0-\mathbf{y}_0)\right) =\mathbf{0}.$$

So a neccesary condition is that the transition matrix has a zero eigenvalue.

Case $\lim_{t\to \infty} |\mathbf{x}_t - \mathbf{y}_t| = 0$

The 'successful coupling' can mean that the distance between the two chains approaches one (if you wait long enough then the coupling occurs almost surely).

Example for a transition matrix

$$\mathbf{P} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\end{bmatrix}$$

We can have the joint process $\mathbf{u}_t,\mathbf{v}_t$ that transitions according to

$$\begin{array}{c|ccccccccc} &1,1 & 2,2 & 3,3 & 1,2 & 1, 3& 2,1& 2,3 & 3,1 &3,2\\ \hline 1,1 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} &\frac{1}{3} & \frac{1}{3} &0&0&0&0\\ 2,2& \frac{1}{3} & \frac{1}{3} & \frac{1}{3}&0&0&\frac{1}{3}&\frac{1}{3}&0&0\\ 3,3& \frac{1}{3} & \frac{1}{3} & \frac{1}{3}&0&0&0&0&\frac{1}{3}&\frac{1}{3}\\ 1,2&0&0&0&\frac{1}{3}& \frac{1}{3}&0&0&0&0\\ 1,3&0&0&0&\frac{1}{3}& \frac{1}{3}&0&0&0&0\\ 2,1&0&0&0&0&0&\frac{1}{3}&\frac{1}{3}&0&0\\ 2,3&0&0&0&0&0&\frac{1}{3}&\frac{1}{3} &0&0\\ 3,1&0&0&0&0&0&0&0&\frac{1}{3}&\frac{1}{3}\\ 3,2&0&0&0&0&0&0&0&\frac{1}{3}&\frac{1}{3}\\ \end{array}$$

This will eventually end up in the states (1,1), (2,2) and (3,3) where $U_t=V_t$ while the marginal distributions of process are the same as $X_t$ and $Y_t$ if we start with $U_0 = X_0$ and $V_0 = Y_0$.

However this is not the same as $$P[U_t \neq V_t] = 0;$$ instead we have $$P[U_t \neq V_t] = 0.75 \left(\frac{2}{3}\right)^t;$$ we have $P[U_0 \neq V_0] = 0.75$ and every step there is a $2/3$ probability that they remain different.

So, with this interpretation of successful coupling, it occurs when either both $\alpha$ and $\beta$ have the same stationary state, or have different stationary states with equal probability, or end up in similar cyclic states at the same time.

It is not easy to express in a simple way all the matrices that fulfill such condition.