More generally speaking, you're asking when the iteration `x_+ = Ax` converges to a fixed point which is an eigenvector of A. This can happen a few different ways. The obvious way is that A has an eigenvector `v` with eigenvalue 1, and all other eigenvalues with magnitude < 1. Then those other components will die out with repeated application of A, leaving only `v` in the limit.
For Markov chains, we can get this exact property from the Perron-Frobenius theorem, which applies to non-negative irreducible matrices. Irreducible means that the transition graph of the Markov chain is strongly connected. If that's the case, then there is a unique eigenvector called the stationary distribution (with eigenvalue 1), and all initial conditions will converge to it.
In case A is not irreducible, you may have different connected components, and the stationary distribution may depend on which component your initial condition is in. Going back to the n x n identity matrix, it has n connected components (it's a completely disconnected graph with all the self-transition probabilities = 1). So every initial condition is stationary, because you can't change anything after the initial step.
Beyond just markov chains, matrices do “move” vectors towards eigenvectors sometimes. If a matrix A has eigenvectors x1 and x2 with eigenvalues r1 and r2, A(x1+x2)= r1x1+r2x2 because matrix multiplication is a linear transformation. If we repeatedly multiply x1+x2 by A, A^n(x1+x2)=r1^nx1+ r2^nx2. Then, if r1>r2, all of the terms are growing exponentially but the contribution of x1 to the result grows exponentially faster than the contribution of x2 so for some large n, A^n(x1+x2) = r1^nx1 + some irrelevant error term.
This means the largest eigenvalues sort of dominate, and you might especially care about eigenvalues of 1, because those mean Ax=x so x is a steady state and if you can write down A as a matrix you can solve for non zero x and learn about the steady state solutions.