= Solution
View every $k$-automorphism $\sigma_i:K\to K$ as an element of the $K$-vector space
$$
\operatorname{Hom}_k(K,K),
$$
where $K$ acts by multiplying output values. If $n=[K:k]$ and $e_1,\ldots,e_n$ is a $k$-basis of $K$, a $k$-linear map is determined freely by its $n$ values on that basis. Thus
$$
\dim_K\operatorname{Hom}_k(K,K)=n.
$$
Part (a) says that the $m$ distinct automorphisms are <linearly independent vectors> in this space, so
$$
\boxed{m\leq[K:k]}.
$$
Now let $k=K^G$, the <fixed field> of the finite automorphism group $G$. Its elements are fixed by every member of $G$, so the members of $G$ are distinct $K^G$-automorphisms of $K$. Applying the inequality gives the <automorphism-count bound for a finite field extension>
$$
\boxed{|G|\leq[K:K^G]}.
$$
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