= 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. Thus
$$
\boxed{\#\operatorname{Fix}(\sigma)=\sum_{i=1}^m A_{ii}=\operatorname{tr}A.}
$$
This also covers an empty <subshift of finite type>, for then no permitted constant <sequence> exists.
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