Do there exist Markov partitions with (nearly) uniform SRB measures? Let $M$ be a compact, finite-dimensional Riemannian manifold, let $T: M \rightarrow M$ be an Anosov diffeomorphism, and let $\mu$ be a Sinai-Ruelle-Bowen (probability) measure. Write $\mathcal{R} = \{ R_1,\dots,R_n \}$ for a Markov partition; write $p_j^{(\mathcal{R})} := \mu(R_j)$.
Question:
Does there ever/always exist $\mathcal{R}$ s.t. $p^{(\mathcal{R})}$ is a nontrivial uniform measure on $\{1,\dots,n\}$? If not, does there ever/always exist a sequence of partitions $\mathcal{R}_m$ s.t. $p^{(\mathcal{R}_m)}$ converges to a uniform measure in some nontrivial sense?
 A: Since (p_1....,p_n) is an eigenvector of the transition matrix, your uniformity assumption is satisfied iff the transition matrix associated to the partition is bistochastic. I guess this is rarely the case.
A: I wanted to elaborate on coudy's original answer, but I also think something is wrong with this argument. In the absence of any other answers I will make the bounty available for an explanation of what goes wrong here...

If $\mathcal{P}$ is a partition of $M$, write $\mathcal{P}_m^\vee := \bigvee_{j=0}^m T^j \mathcal{P}$. (Because $T$ preserves the SRB measure $\mu$ we don't have to consider, e.g., $j<0$ terms.)
A Markov partition $\mathcal{R}$ is generating, so the supremum over partitions in the Kolmogorov-Sinai entropy is realized by it: the KS entropy is $h_\mu^{KS}(T) = -\lim_m m^{-1}\sum_{R \in \mathcal{R}_m^\vee} \mu(R) \log \mu(R)$. Meanwhile the topological entropy is $h(T) = \lim_m m^{-1} \log \# \mathcal{R}_m^\vee$. 
So if $\mu$ is also the measure of maximal entropy (as is the case, e.g. when $T$ is a hyperbolic toral automorphism or the Poincaré map for a geodesic flow on a surface of negative curvature), then $h_\mu^{KS}(T) = h(T)$ and in the limit we have that $-\sum_{R \in \mathcal{R}_m^\vee} \mu(R) \log \mu(R) \sim \log \# \mathcal{R}_m^\vee$, so that $\mu$ is asymptotically uniform on $\mathcal{R}_m^\vee$.
