Linear maps of polynomial roots over finite fields Take an arbitrary degree $n$ monic irreducible polynomial $f(x)$ over $\mathbb{F}_p$ with roots
$$\alpha, \alpha^p,\alpha^{p^2},\dots,\alpha^{p^{n-1}}$$
in $\mathbb{F}_p[x] / \langle f(x) \rangle$. Now, consider another degree $n$ monic irreducible $g(x)$ over $\mathbb{F}_p$. Is there a linear map which sends the roots of $f$ to the roots of $g$ in $\mathbb{F}_{p^n}\cong \mathbb{F}_p[x] / \langle f(x) \rangle$ (viewed as a vector space)?
Edit: More specifically, if $\alpha,\dots,\alpha^{p^{n-1}}$ are not a normal basis for $\mathbb{F}_{p^n}$, what (if anything) can be said?
 A: I will use this example: Constructing an explicit isomorphism between finite extensions of finite fields
Let $p=2$, $f(x)  = x^3+x+1$, $g(x)=x^3+x^2+1$ so that $L\simeq \mathbb{F}_2[x]/(x^3+x+1)$ and $L'\simeq \mathbb{F}_2[x]/(x^3+x^2+1)$ are isomorphic as fields. Let $\alpha$ be a root of $f(x)$ in $L$, and $\beta$ be a root of $g(x)$ in $L'$. 
Since $f$ does not have $x^2$ term, we see that $\alpha^4+\alpha^2+\alpha =0$. 
For the roots of $g$, we have $\beta^4+\beta^2+\beta = 1 \in \mathbb{F}_2$. 
We assume that  $T:L\rightarrow L'$ is  a  $\mathbb{F}_2$-linear map that maps roots of $f$ to roots of $g$, i. e. 
$$
T(\{\alpha,\alpha^2,\alpha^4\})\subseteq \{\beta,\beta^2,\beta^4\}.
$$
Now, we consider $\alpha^4=\alpha^2+\alpha$. So, $T(\alpha^4)=T(\alpha^2)+T(\alpha)$.
If $T(\alpha)=T(\alpha^2)$, then we have $T(\alpha^4)=0$.  This is not possible. 
Then we must have $T(\alpha)\neq T(\alpha^2)$. We check the three possibilities: 
Case 1: $T(\{\alpha,\alpha^2\})=\{\beta,\beta^2\}$. 
Since $T(\alpha^4) = \beta^2+\beta$ and $\beta^4=\beta^2+\beta+1 $,  the element $\beta^2+\beta$ is not a root of $g$, this case is impossible. 
Case 2: $T(\{\alpha,\alpha^2\})=\{\beta,\beta^4\}$. 
Since $T(\alpha^4)= \beta^4+\beta  = \beta^2+1$ is not a root of $g$, it is not possible. 
Case 3: $T(\{\alpha,\alpha^2\})=\{\beta^2,\beta^4\}$.
Since $T(\alpha^4) = \beta^4+\beta^2 =  \beta+1$ is not a root of $g$, it is not possible. 
