= Solution
Use the same intended <weak-star topology> as in (b), and assume $C\ne\varnothing$. The induced operator is $T=S^*$, so $T\phi(x)=\phi(Sx)$; it is weak-star continuous because evaluation after applying $T$ is still evaluation at a <vector> of $E$.
Choose $\phi\in C$ and form the averages
$$
\phi_n=\frac1n\sum_{j=0}^{n-1}T^j\phi.
$$
Invariance and convexity put every $\phi_n$ in $C$. Weak-star closedness and (b) make $C$ compact, so some subsequence converges to $\psi\in C$. Telescoping gives
$$
T\phi_n-\phi_n=\frac{T^n\phi-\phi}{n},\qquad
\|T\phi_n-\phi_n\|\leq\frac2n,
$$
because both orbit points lie in the dual <unit ball>. Weak-star <continuity> now gives $T\psi-\psi=0$. Hence $\boxed{T\text{ has a fixed point in }C.}$ This is the <weak-star fixed point theorem for an adjoint operator>; it does not require $T$ to be a contraction on the entire <dual space>.
If “weakly closed” is interpreted literally as closed for $\sigma(E^*,E^{**})$, the result can fail. Take $E=c_0$, let $S$ be the left shift and $T=S^*$ the right shift on $\ell^1$. The set
$$
C=\{x\in\ell^1:x_j\geq0,\ \sum_jx_j=1\}
$$
is a nonempty weakly closed convex subset of the <unit ball>: its defining coordinate inequalities and sum equality are weakly closed. It is invariant under the right shift. But a fixed <vector> of that shift has first coordinate zero and then every coordinate zero, so none belongs to $C$. This supplies a counterexample to the literal topology and explains why the weak-star qualification is essential.
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