searching an explicit isomorphism of finite fields Since all the finite field of $p^n$ elements are the splitting field of the separable polynomial $x^{p^n}-x$, all of them are isomphic.
In particular if $f_1(x),f_2(x)$ are irreducible polynomials over $\mathbb{F}_p[x]$ of the same degree. Then:  $$ \mathbb{F}_p[x]/(f_1(x)) \cong \mathbb{F}_p[x]/(f_2(x))$$
But I want to find an explicit isomorphism. I don't know if it's always possible. But the following could be useful.
Let's consider $f_2$ as a polynomial in $(\mathbb{F}_p[x]/(f_1(x)))[y]$. If $\gamma(x) $ is a root of $f_2(y)$ (a root on $\mathbb{F}_p[x]/(f_1(x))$). Then the following map is an isomorphism:
$$\mathbb{F}_p[x]/(f_2(x)): \to \mathbb{F}_p[x]/(f_1(x)) $$
$$x\to \gamma(x)$$
My question if there are techniques to find that $\gamma(x)$. 
For example if $f_1 = x^4+x^3+1 , f_2 = x^4+x+1 $ are over $\mathbb{F}_2$ then $\gamma(x)=x^3+x^2 $ it's a root.
If the solution of the general case it's not possible (or unsolved) or too difficult, I want to know at least this particular case :/
I want to compute it in the case $ f_1= x^2+2x+2 , f_2 = x^2+x+3 $ over  $\mathbb{F}_7$
 A: In the field ${\mathbb F}_7[w]$ where  $w^2 + 2 w + 2 = 0$, you want to find $\alpha$ such that $\alpha^2 + \alpha + 3 = 0$.  It must be of the form $a + b w$ with $a, b \in {\mathbb F}_7$.  Well, $(a+b w)^2 + a + b w + 3 = {a}^{2}+3+a-2\,{b}^{2}+ \left( b+2\,ab-2\,{b}^{2} \right) w$, so we want $ {a}^{2}+3+a-2\,{b}^{2} = 0$ and $ b+2\,ab-2\,{b}^{2} = 0$ mod 7.
The solutions are $a=2,b=6$ and $a=4,b=1$.  
A: You can fctorize $f_2(y)\in\mathbb{F}_p[x]/(f_1(x))$ into linear facrors using factorization algorithms like  Cantor–Zassenhaus algorithm or
Berlekamp's algorithm . I think there are no easier way to do this in the general case.$ {}  {} {} $
A: At least for the case $n=2$, I believe the following works.
Let $q(x)$ and $r(x)$ be your irreducible polynomials. Without loss of generality, they're monic. We can complete the square and write them as $q(x) = (x+a)^2 + b$ and $r(x) = (x+c)^2 + d$. 
Choose an automorphism of $\mathbb{F}_{p}$ called $\phi$ such that $\phi(b) = d$ (namely, multiplication of elements in $\mathbb{F}_p$ by $db^{-1}$). This automorphism induces an automorphism $\phi^\star$ of $\mathbb{F}_{p} [x]$ given by $$\phi^\star (a_n x^n + \cdots + a_0) = \phi(a_n) x^n + \cdots + \phi(a_0)$$
Let $k = c- \phi(a)$.
Then, the map $a + (q(x)) \mapsto \phi^\star (a) (x+k) + (r(x))$ is an isomorphism between $\mathbb{F}_{p}[x]/((q(x))$ and $\mathbb{F}_{p}[x]/((r(x))$.
