Galois group of $x^4-x^2-3$ on $\mathbb{F}_5$ As stated in the title, the problem concerns 
$$\text{Gal}_{\mathbb{F}_5}(x^4-x^2-3)$$
I managed to prove that
$$\text{Gal}_{\mathbb{F}_5}(x^4-x^2-3)\simeq \mathbb{Z}_4$$
but I'm not sure if my reasoning is correct, and I'd like a verification:
First, we note that the polynomial is irreducible over the base field, since its roots are $\pm\sqrt{3\pm3\sqrt{3}}\notin\mathbb{F}_5$ (which follows from the fact that $\sqrt{3}\notin \mathbb{F}_5$, which I proved using the legendre symbol), and thus cannot be written as a product of a linear factor and a cubic one; moreover, since no product of two of its linear factor in $\overline{\mathbb{F}}_5[X]$
is in $\mathbb{F}_5[X]$ it cannot be written
as the product of two quadratic polynomials.
Now, we want to compute the order of the Galois group of the polynomial. Since the
polynomial is irreducible and its derivative is not zero, the splitting field is a separable extension on $\mathbb{F}_5$, and so
$$\text{ord}(\text{Gal})=\left[\mathbb{F}_5\left(\sqrt{3+3\sqrt{3}},\sqrt{3-3\sqrt{3}}\right):\mathbb{F}_5\right]$$
Now we note that $\sqrt{3-3\sqrt{3}}=\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3+3\sqrt{3}}}$, and so we obtain
$$\text{ord}(\text{Gal})=\left[\mathbb{F}_5\left(\sqrt{3+3\sqrt{3}},\sqrt{3-3\sqrt{3}}\right):\mathbb{F}_5\right]=\left[\mathbb{F}_5\left(\sqrt{3+3\sqrt{3}}\right):\mathbb{F}_5\right]=4$$
Thus, we need to find a subgroup of order $4$ of the permutation group of $\{\alpha,-\alpha,\beta,-\beta\}$ (where $\alpha=\sqrt{3+3\sqrt{3}}$ and $\beta$ is the other solution)
it can easily be seen that there are only $8$ permutations that are also field morphism (i.e. such that $\sigma(-\alpha)=-\sigma(\alpha);\sigma(-\beta)=-\sigma(\beta)$). Between those, the permutation 
$$\sigma: \begin{cases}\alpha\to -\beta\\
\beta\to \alpha \end{cases}$$
creates a group of order $4$. Thus, $\text{Gal}$ is a cyclic group of order four, i.e.
$$\text{Gal}_{\mathbb{F}_5}(x^4-x^2-3)\simeq \mathbb{Z}_4$$
I have two problems with these proof.
The first one is that I'm not sure on how to prove that, aside from being a permutation of the roots, $\sigma$ is effectively a field automorphism (actually, I'm not sure of that: since $\beta=\frac{1}{(\alpha^2-3)\alpha}$, maybe there's some other conditions that I have to impose on the permutation ).
The second one is that I wonder if there's a shorter path to the solution, a more "programmatic" approach then guessing the correct root permutation.
 A: The key argument to shorten your proof is the following result: if $L$ is a finite extension of a finite field $K$ then $L/K$ is Galois and its Galois group is cyclic of order $[L:K]$. 
The proof is as follows: note that $L$ and $K$ have the same characteristic $p$, write $q=|K|$ and $r=|L|=q^d < \infty$. 
Let $F$ be the map $L \rightarrow L$ such that $F(x)=x^q$. Then $F$ is additive (characteristic and properties of binomial coefficients), multiplicative (trivial), injective and $F_{|K}=id$, so (finite cardinality) it is an element of the group $G$ of $K$-automorphisms of $L$. 
We know that $G$ has cardinality at most $d$ with equality iff $L/K$ is Galois. 
But if $1 \leq k < d$, $F^k: x \longmapsto x^{q^k}$ which is not the identity on $L$: otherwise the nonzero polynomial $X^{q^k}-X$ with degree $<|L|$ vanishes on all of $L$, a contradiction. 
So $id,F,F^2,\ldots,F^{d-1}$ are pairwise distinct elements of $G$ so $|G| \geq d$, so $|G|=d$ and $G$ generated by $F$. 
The application of the result to your problem is spectacular: let $P$ be a monic irreducible polynomial of degree $n$ over a finite field $F$. Then its Galois group is cyclic of order $n$. 
Indeed, let $G$ be the Galois group. $P$ is separable (finite fields are perfect) so it has $n$ pairwise distinct roots. $G$ acts transitively on these roots so $|G|$ must be divisible by $n$. 
Now, let $E$ is a field extension of $F$ of dimension $n$ containing one root of $P$ (eg $E=F[X]/(P)$). Then $E/F$ is normal so $P$ splits in $E$. So the splitting field $K$ of $P$ is a subextension of $E$ so $|G|=[K:F]$ isn’t greater than $n=[E:K]$. So $|G|=n$ and by the above $G$ is cyclic. 
