 Solution

A fixed point of the left shift satisfies $x_{k+1}=x_k$ for every $k\in \mathbb{Z}$, so its entire two-sided sequence is constant. The constant sequence with value $i$ belongs to the subshift of finite type precisely when $A_{ii}=1$: this is exactly the requirement that its repeated transition be allowed by the symbolic transition matrix. There is one such fixed point for each diagonal entry equal to one. ThusThis also covers an empty subshift of finite type, for then no permitted constant sequence exists.