Group-Isomorphism problem I want to find an group-isomorphism $$ \psi : (\mathbb{Z}/8\mathbb{Z},+) \longrightarrow \mathbb{F}_9^\times $$
which should be used to multiply elements in $\mathbb{F}_9$ or to find the inverse element in an easy way.
 A: As Seth and Pedro indicated the existence of such an isomorphism follows from (and is equivalent to) the cyclicity of the multiplicative group $\Bbb{F}_9^*$. To exhibit an explicit isomorphism you need to specify a construction of $\Bbb{F}_9$ and find a generator of the multiplicative group (aka a primitive element).
Elements of order $8$ are zeros of the cyclotomic polynomial $\phi_8(x)=x^4+1$. As we are working modulo $3$, we can factor it as follows
$$
\begin{aligned}
x^4+1&=x^4+4=(x^4+4x^2+4)-4x^2\\
&=(x^2+2)^2-(2x)^2=(x^2+2-2x)(x^2+2+2x)\\
&=(x^2-x-1)(x^2+x-1).
\end{aligned}
$$
So, if we identify $\Bbb{F}_9$ with $\Bbb{Z}_3[x]/\langle x^2-x-1\rangle$, then the coset
$\alpha=x+\langle x^2-x-1\rangle$ will be a generator.
The desired isomorphism $\psi:(\Bbb{Z}_8,+)\to (\Bbb{F}_9^*,\cdot)$ is given by
$$
\psi(\overline{n})=\alpha^n
$$
for all $n=0,1,\ldots,7$. This comes to the following:
$$
\begin{array}{c|ccc|c}
n&\psi(\overline{n})&&n&\psi(\overline{n})\\
\hline
0&1&&1&\alpha\\
2&\alpha^2=\alpha+1&&3&\alpha^3=\alpha^2+\alpha=2\alpha+1\\
4&\alpha^4=2&&5&\alpha^5=2\alpha\\
6&\alpha^6=2\alpha^2=2\alpha+2&&7&\alpha^7=\alpha+2
\end{array}
$$
Here I used the equation $\alpha^2=\alpha+1$ repeatedly. Note also that $\alpha^4=2=-1$, so
$\alpha^{4+j}=-\alpha^j$ for all $j$. As a final check we do that
$$
\alpha^8=\alpha\cdot\alpha^7=\alpha(\alpha+2)=\alpha^2+2\alpha=3\alpha+1=1
$$
as it should.
The way to use this in finding products and inverses is to treat the inverse of $\psi$ as a logarithm, i.e. a mapping that transforms multiplication in $\Bbb{F}_9$ into addition in $\Bbb{Z}_8$. For example (read the above table from right to left)
$$
(\alpha+2)(\alpha+1)=\alpha^7\cdot\alpha^2=\alpha^9=\alpha^{8+1}=\alpha,
$$
and
$$
(\alpha+1)^{-1}=(\alpha^2)^{-1}=\alpha^{-2}=\alpha^{8-2}=\alpha^6=2\alpha+2.
$$
You mentioned that you had found $x+2=x-1$ and $2x+1$ to be generators of $\Bbb{F}_9^*$.
That may be true, but to make that meaningful you need to specify the minimal polynomial of $x$. Above I used the minimal polynomial $x^2-x-1$, and $\alpha$ was one of its zeros ($\alpha^3=2\alpha+1$ being the other. If a cyclic group of order $8$ is generated by an element $c$, then $c^m$ is also a generator iff $\gcd(m,8)=1$. So here any of $\alpha$, $\alpha^3$, $\alpha^5$ and $\alpha^7$ is a generator. Not surprisingly those are the zeros of $x^4+1$ in this copy of $\Bbb{F}_9$.

Changing the irreducible polynomial always changes the scene. Your copy of $\Bbb{F}_9$ is $\Bbb{Z}_3[x]/\langle x^2+1\rangle$. Because $x^2+1\mid x^4-1$ in this case $x+\langle x^2+1\rangle$ will not work as a generator. If we let $\beta=x+\langle x^2+1\rangle$, then 
for $\alpha=\beta+2$ (that you might call $x+2$ even though that abuses notation slightly)
we get
$$
0+\langle x^2+1\rangle =(x^2+1)+\langle x^2+1\rangle=(\alpha+1)^2+1=\alpha^2+2\alpha+2=\alpha^2-\alpha-1.
$$
Thus $x^2-x-1$ is a minimal polynomial of $\alpha$. In terms of $\beta$ the above the homomorphism looks like
$$
\begin{array}{c|ccc|c}
n&\psi(\overline{n})&&n&\psi(\overline{n})\\
\hline
0&1&&1&\beta+1\\
2&(\beta+1)^2=2\beta&&3&(\beta+1)^3=2\beta^2+2\beta=2\beta+1\\
4&(\beta+1)^4=2&&5&(\beta+1)^5=2\beta+2\\
6&(\beta+1)^6=\beta&&7&(\beta+1)^7=\beta+2
\end{array}
$$
All the elements on the right column ($\beta+1,2\beta+1,2\beta+2,\beta+2$) can take the role of the generator. Thus you get four different such isomorphisms $\psi$. Each will work equally well as the inverse of a discrete logarithm.
A: It suffices you show the group of units of any finite field is cyclic. In fact, any finite subgroup of the group of units of any field is cyclic: $x^d=1$ has at most $d$ solutions in any finite subgroups of $F^\times$, which means any finite subgroups has at most one cyclic subgroup of order $d$.
