How to find a explicit field morphism I want to find a explicit morphism from $K_1$ to $K_2$, where

$$K_1=\mathbb F_2[x]/(x^3+x+1)\mbox{ and }K_2=\mathbb F_2[x]/(x^3+x^2+1).$$

I've found that it must exist because these polynomials are irreducible, hence these fields have $2^3$ and $2^3$ elements and $3\mid 3$.

But how could I find the explicit morphism?
Since the multiplicative group $\mathbb F_8^*$ is the cyclic group of order $7$, any nontrivial element generates the full group. For example, in $K_1$, the group is $\{1,x,x^2, \ldots x^6\}$.
Next, we know that the Frobenius map $x \mapsto x^2$ is a field automorphism, so that $x,x^2,x^4$ must have the same minimal polynomial, and likewise with $x^3,x^5,x^6$.
Since the minimal polynomial of $x$ is $X^3+X+1$, this means that $x^3,x^5,x^6$ are the three roots of $X^3+X^2+1$ in $K_1$. Thus sending the $x$ of $K_2$ to any of those will work.

For example the map $K_2 \to K_1$ given by $x \mapsto x^{-1} (= x^6)$ is an isomorphism (and now we notice that the second polynomial was the first polynomial with the coefficients reversed)
Hint: Both fields are isomorphic to $\mathbb{F}_{2^{3}}$ and explicitly
the elements in both of them are of the form $[a+bx+cx^{2}]$ (by
the Euclidean algorithm) (where $a,b,c\in\mathbb{F}_{2})$.

So any homomorphism $\mathbb{F}_{8}\to\mathbb{F}_{8}$ will be a good
answer - given that you have explicitly isomorphisms between $\mathbb{F}_{8}$
to those fields. such isomorphism can be found by finding the multiplication
table of both field and comparing it with the one of $\mathbb{F}_{8}$.