 Solution

Define $\vartheta (x)=x^p$ on all of $K$. The binomial coefficients between the endpoints are divisible by $p$, so $(x+y{)}^p=x^p+y^p$; also $(xy{)}^p=x^p{y}^p$ and $1^p=1$. Thus this is a field homomorphism. Its kernel is zero, and an injective self-map of the finite set $K$ is surjective. Hence $\boxed{\vartheta \text{ is an automorphism}}$, the Frobenius automorphism, with the required image of $\lambda$. It fixes $F$ by the preceding prime-subfield argument.