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)