I am looking for an explicit description of the additive, multiplicative group structure and automorphism group of the field with 729 elements.
Sorry for not being clear. I have no clue what does "explicit description" mean, but I am guessing that the underlying group might be cyclic, but I do not know how to give a proof.
For its automorphism group, What I have tried: 1). Comparing this with the Galois group of this field over $F_3$ which is cyclic. 2). I also know that Aut(Z_3） = Z_2, which is, in general, true for or prime fields. That is why I am guessing that the underlying additive group should be cyclic, then by the same method, I can show that the automorphism group is a cyclic group of order 728 probably.
A field $F$ with $729$ elements is a vector space of dimension $6$ over $\mathbb F_3$ because $729=3^6$. Therefore, the additive group of $F$ is isomorphic to $\mathbb F_3^6$, the cartesian product of six copies of $\mathbb F_3$.
The multiplicative group of every finite field is cyclic. Therefore, the multiplicative group of $F$ is cyclic of order $728$.
$F$ is a simple extension of $\mathbb F_3$ generated by any root $\theta$ of any polynomial of degree $6$ that is irreducible mod $3$. The simplest one if $x^6+x+2$. An automorphism of $F$ must send $\theta$ to another root of the same polynomial. Therefore, the automorphism group of $F$ has order $6$. You just have to decide whether it is $C_6$ or $S_3$.
Alternatively, $F$ is a splitting filed (of $x^{729}-x$ for instance) and so is a Galois extension of $\mathbb F_3$. Since its only proper subfields are $\mathbb F_{3^2}$ and $\mathbb F_{3^3}$, the Galois group must be $C_6$.