= Solution
The <polynomial> $X^m-1$ is separable, since its <derivative> $mX^{m-1}$ has no common root with it when the <characteristic> does not divide $m$. Its roots form a cyclic multiplicative group of order $m$. For completeness, any finite subgroup of a field's multiplicative group is cyclic: if its exponent is $e$, all its elements are roots of $X^e-1$, so its size is at most $e$, while the elementary decomposition of a finite abelian group gives an element of order $e$.
Choose a primitive root $\zeta$. Then $L=K(\zeta)$, and every <field automorphism> sends $\zeta$ to $\zeta^a$ for a unique unit $a$ modulo $m$. Composition multiplies the exponents, and fixing $\zeta$ fixes all of $L$. This is the required injective homomorphism
$$
\boxed{\operatorname{Gal}(L/K)\hookrightarrow(\mathbb Z/m\mathbb Z)^*.}
$$
For $K=\mathbb F_q$, the least extension $\mathbb F_{q^d}$ containing a primitive $m$th root is characterized by $m\mid q^d-1$, since its multiplicative group is cyclic of order $q^d-1$. Thus $[L:K]$ is the multiplicative order of $q$ modulo $m$. The successive powers of $4$ modulo $11$ are $4,5,9,3,1$, so
$$
\boxed{[\mathbb F_4(\zeta_{11}):\mathbb F_4]=5.}
$$
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