Suppose we have two fields $F_1$ and $F_2$ of order $9$ where both groups of units are cyclic, i.e. $$F_1=\{0\}\cup\{\alpha^i\,|\,0\leq i\leq 7\},\qquad F_2=\{0\}\cup\{\beta^i\,|\,0\leq i\leq 7\}$$ for some $\alpha\in F_1$ and $\beta\in F_2$. We know that those two fields have to be isomorphic as they have the same number of elements. A concrete isomorphism would be given by $\varphi: F_1\to F_2$ where $\varphi(\alpha):=\beta$. It is easy to see that $\varphi(xy)=\varphi(x)\varphi(y)$ for every $x,y\in F_1$ due to the representation of the fields as powers of the generators $\alpha$ and $\beta$; the fact that $\varphi$ is bijective follows immediately as well.
It is not clear to me, however, why $\varphi(x+y)=\varphi(x)+\varphi(y)$ for every $x,y\in F_1$. Is there any elementary way to see this?
Edit: As metamorphy noted in the answer below, $\varphi$ is not generally an isomorphism. Is there any way, then, to choose another suitable generator $\beta'\in F_2^\times$ such that $\varphi$ is actually an isomorphism?
$\varphi(\alpha):=\beta$ (i.e. $\varphi(\alpha^i):=\beta^i$) does not necessarily define an isomorphism.
The automorphisms of $\mathbb{F}_q$, for $q=p^n$ and $p$ prime, are $x\mapsto x^{p^k}$ for $0\leqslant k<n$. In particular, for $q=9$, the only nontrivial automorphism is $x\mapsto x^3$. Hence, for $F_1=F_2=\mathbb{F}_9$ and $\beta=\alpha^5$ (say), the corresponding $\varphi$ is not an isomorphism.
As for the final subquestion, the choice of $\beta'$ is determined by the action of $+$ (in $F_1$ and $F_2$) or, as noted in the comments, by the action of $x\mapsto x+1$. For $\mathbb{F}_9$, the (only) two possibilities are:
\begin{array}{c|c|c|c|c|c|c|c|c|} x & 0 & 1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6 & \alpha^7 \\ \hline (x+1)_1 & 1 & \alpha^4 & \alpha^7 & \alpha^3 & \alpha^5 & 0 & \alpha^2 & \alpha & \alpha^6 \\ \hline (x+1)_2 & 1 & \alpha^4 & \alpha^2 & \alpha^7 & \alpha^6 & 0 & \alpha^3 & \alpha^5 & \alpha \\ \hline \end{array}