= Solution
Fix $z_0\in S^1$, and let $\delta_{z_0}\in C(S^1)^*$ be the <point evaluation functional> $\delta_{z_0}(f)=f(z_0)$. Define orbit averages in the dual space by
$$
\mu_n=\frac1n\sum_{j=0}^{n-1}(T^*)^j\delta_{z_0}.
$$
Each $\mu_n$ has norm at most one and satisfies $\mu_n(1_{S^1})=1$. The space $C(S^1)$ is separable, so part (a) gives a subsequence converging pointwise to some $\mu\in C(S^1)^*$. In particular,
$$
\mu(1_{S^1})=1.
$$
For every $f\in C(S^1)$, the <Cesaro average> telescopes at the endpoints:
$$
\mu_n(Tf)-\mu_n(f)
=\frac{\delta_{z_0}(T^nf)-\delta_{z_0}(f)}n,
$$
whose absolute value is at most $2\|f\|_\infty/n$. Passing to the convergent subsequence proves
$$
\boxed{\mu(Tf)=\mu(f)\quad\text{for every }f\in C(S^1)}.
$$
Thus $\mu$ is an <invariant functional of a composition operator>.
It need not be unique when $T$ is not the <identity operator>. For the nonidentity <homeomorphism> $\tau(z)=\overline z$, both $1$ and $-1$ are fixed. Hence the two distinct point evaluations
$$
\delta_1(f)=f(1),
\qquad
\delta_{-1}(f)=f(-1)
$$
have value one on $1_{S^1}$ and are invariant under $T$. Therefore \b[nonidentity does not imply uniqueness].
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