Let $P_0(x)$ and $P_1(x)$ be two given distinct irreducible binary polynomials of same degree $k$. For both $P_i(x)$, the set $\mathbb F_{2^k}$ of binary polynomial of degree less than $d$, together with polynomial addition, and polynomial multiplication followed by reduction modulo $P_i$, form a field. It's well-known these fields are isomorphic (see e.g. this question).
Is that isomorphism unique? How can we express it (or one) using $P_0(x)$ and $P_1(x)$?
That is, is there a unique permutation $g$ on the set $\mathbb F_{2^k}$ such that $$\begin{align}\forall A(x)\in\mathbb F_{2^k},\,\forall B(x)\in\mathbb F_{2^k},\ &g\big(A(x)+ B(x)\big)=g\big(A(x)\big)+ g\big(B(x)\big)\\ &g\big(A(x)\cdot B(x)\bmod P_0(x)\big)=g\big(A(x)\big)\cdot g\big(B(x)\big)\bmod P_1(x) \end{align}$$ and how do we express it (or one)?
I see that $g(0)=0$, $g(1)=1$, and it's enough to know $g(x)$ to fully define $g$. But I fail to prove that $g$ is unique, or find what $g(x)$ can be. I've tried to suppose there are two distinct $g$ and $g'$, define $h=g'\circ g^{-1}$, state there is a polynomial $D(x)$ with $h(D(x))\ne D(x)$ and come to a contradiction, but nothing happens.
$P_1(T)\in \Bbb{F}_2[T]$ has a root $a$ in the field $\Bbb{F}_2[x]/(P_0(x))$, the other ones are $a^{p^m}$, and the isomorphisms $$\Bbb{F}_2[y]/(P_1(y))\to \Bbb{F}_2[x]/(P_0(x))$$ send $y$ to $a^{p^m}$ for some $m$ ($k$ possible choices).