Consider a finite set of dimension $N$, and let $P$ denote a $N\times N$ row-stochastic matrix of transition probabilities on this set. For each $P$, let $M\left(P\right)$ denote the set of its invariant distributions:$$M\left(P\right)=\left\{ \mu\in\Delta^{N-1}:\,\mu'P=\mu\right\} .$$ Does $M$ admit a continuous selection? That is, does there exist a map $\mu\left(P\right)$ continuous in $P$ such that $\mu\left(P\right)'P=\mu\left(P\right)$ for each transition matrix $P$?
If not, can we find such a map at least on the interior of the set of transition matrices?
Below are some thoughts, but feel free to follow other routes:
The Brouwer's fixed-point theorem ensures that $M\left(P\right)$ is non-empty for each $P$ (indeed, for a fixed P, the map $\mu\mapsto\mu'P$ is a continuous on $\Delta^{N-1}$). Moreover, $M$ is closed and convex-valued. If it were lower-hemicontinuous, we could use Michael’s Selection Theorem. So the question could be reduced to “is $M$ lower-hemicontinuous?”.
Let $P(a,b)=\pmatrix{1-a&a\\b&1-b}$, whose unique stationary distribution (if $0<a,b<1$) is $\mu(a,b)=(b/(a+b),a/(a+b))$. It is easy to check that the rational function $\mu(a,b)$ is discontinuous at $(0,0)$: both $(1,0)=\lim_{\to0}\mu(b^2,b)$ and $(0,1)=\lim_{a\to0}\mu(a,a^2)$ are limit points of $\mu(a,b)$ as $(a,b)\to(0,0)$.
So the answer is no: a continuous selection on the set of all transition matrices does not exist.
If you restrict to the interior of the set of transition matrices, the Perron Frobenius theorem tells you that there is a unique stationary distribution. This distribution solves a full rank system of affine equations in the coordinates of your $\Delta^{N-1}$ (as in the example above); the coefficients in this equation system depend continuously on the transition matrix entries, and hence so does their solution $\mu$.