It seems to me that $g(t)$ and $g'(t)$ are reciprocals of each other:
This does mean that if one of them has a zero in some field, then so does the other.
However, you seem to claim that the zeros coincide (you denote the common zero by $\alpha$). This is false in general. It could be true in some particular case, if the minimal polynomial of $\alpha$ should be a factor of both $g$ and $g'$. That is not the case here, as a quick run of Euclid's algorithm shows that the two polynomials have no common factors.
[Edit] A correct claim is that if $g(\alpha)=0$, then $g'(1/\alpha)=0$. As $g(t)$ is irreducible in $F_2[t]$ so is $g'(t)$. They both have eleven zeros in the field $GF(2048)$, but none of them are common because $\gcd(g(t),g'(t))=1$. [/Edit]