In this proof (picture below), the writer defines a map $\theta$, and prove it to be homomorphism and injective. Thus Gal$(F(w)/F)$ is isomorphic to a subgroup of ${Z_n^*}$. Why it cannot be 'isomorphic to the whole ${Z_n^*}$?
I imitate the proof and do it in, for example, $F=F_2$. $w$ is a primitive root of $x^7-1$. Then $\sigma(w)=w^t$, where $t=1,2,3,4,5,6$ are all automorphism. And the map $\theta$: $\sigma\rightarrow t+7Z$. So it seems to be surjection from Gal$(F(w)/F)$ to ${Z_n^*}$.
But $x^7-1=(x-1)(x^3+x+1)(x^3+x^2+1)$. So $[F(w):F]=3$. This contradicts to $|Gal(F(w)/F)|=6$. So what's the problem???
Once you choose a primitive root $w$, any automorphism of $F(w)$ sends $w$ to another root of the minimal polynomial of $w$ over $F$.
Consequently, you do not obtain ALL the possible primitive roots of $1$. In others words, since $X^7-1$ is not irreducible, the Galois group does not act transitively on the roots.
To be more explicit, in your example, say that $w$ is a root of $X^3+X+1$. What are the other roots ? Well, we have : $w(w^6+w^2+1)=w^7+w^3+w=w^3+w+1=0$, so $w^2$ is another root. The last one is $w^4$, since $w^3(w^{12}+w^4+1)=w^{15}+w^7+w^3=w^3+w+1=0$.
Hence, the map $\sigma$ sending for example $w$ to $w^3$ is NOT an automorphism of $F(w)$. In fact, it is not even well defined: $\sigma(0)=0=\sigma(w^2+w+1)=w^4+w^2+1=w(-1-w)+w^2+1=-w+1\neq 0$.