This answer produces $I$ somewhat indirectly. So it actually *does not* what the OP asked for.

The decomposition group $D$, which is the Galois group of $h(x)$ over $\mathbb Q_2$, can be computed as follows: Using resultants, one sees that the minimal polynomial over $\mathbb Q$ of the difference of two distinct roots of $h(x)$ is \begin{equation}
H(y)=y^{20}+6y^{18}+21y^{16}+46y^{14}-116y^{12}+694y^{10}+1837y^8-1810y^6-1527y^4+8560y^2+9584.
\end{equation}
From the factorization of $h(x)$ over $\mathbb Q_2$ we know that $D$ is a transitive subgroup of $S_4$. Over $\mathbb Q_2$ the polynomial $H(y)$ factors into irreducibles of degrees $4,4,4,8$. The only transitive subgroup of $S_4$ which has these orbit lengths on the $20$ pairs of distinct elements of $\{1,2,3,4,5\}$ is the dihedral group of order $8$.

As Will Sawin already remarked, the inertia group $I$ is a subgroup of $S_2\times S_2$. From the factorization of $h(x)$ over $\mathbb F_2$ we furthermore see that $D/I$ has order $2$. So actually $I=S_2\times S_2$.

There is a [paper][1] by Sybilla Beckmann which describes a method to compute inertia groups under certain restrictions. Her theorem does not apply here. The paper finishes with some remarks about how to possibly extend the methods.


  [1]: http://dx.doi.org/10.1006/jabr.1994.1067