= Solution
\b[Yes.] A <cardinal fixed point of the singular cardinal enumeration> is obtained by countable iteration. Start with $\kappa_0=\aleph_0$ and put
$$
\kappa_{n+1}=\sigma_{\kappa_n}.
$$
The <singular cardinal enumeration> is strictly increasing and satisfies $\sigma_\alpha\geq\alpha$ for every <ordinal> $\alpha$; the latter follows by <transfinite induction> for any strictly increasing ordinal-valued enumeration. If equality occurs at some $\kappa_n$, that <cardinal number> is already a witness. Otherwise the sequence is strictly increasing. Put $\kappa=\sup_{n<\omega}\kappa_n$. This is an uncountable <singular cardinal> of <cofinality> $\omega$.
For every $\alpha<\kappa$, some $n$ has $\alpha<\kappa_n$, and therefore
$$
\sigma_\alpha<\sigma_{\kappa_n}=\kappa_{n+1}<\kappa.
$$
Also $\sup_n\sigma_{\kappa_n}=\kappa$, so $\sup_{\alpha<\kappa}\sigma_\alpha=\kappa$. At a <limit ordinal> index, if the supremum of all preceding enumerated cardinals is itself singular, it is exactly the next member: every smaller singular cardinal already has a preceding index. Thus
$$
\boxed{\sigma_\kappa=\kappa}.
$$
This argument uses continuity only at a singular supremum; the enumeration need not be continuous at a <weakly inaccessible cardinal>.
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