 Solution

For nonzero $x\in K$, Lagrange's theorem in $K^{\times }$ gives $x^{p^n-1}=1$; zero satisfies $x^{p^n}=x$ too. Thus ${\vartheta }^n$ is the identity, and (iii) excludes every smaller positive order. Hence $\boxed{|⟨\vartheta ⟩|=n}$.

Any field automorphism of $K$ fixes $F$ and is determined by its value on $\lambda$. That value must be a root of the degree-$n$ minimal polynomial, so there are at most $n$ automorphisms. The $n$ powers of $\vartheta$ already attain this bound. Consequently $\boxed{\mathrm{Aut}(K)=⟨\vartheta ⟩\cong C_n}$.