**Short answer:** $G = H$ is the group of conformal symplectic linear maps. What follows is a proof of this (which I've simplified slightly from what I originally wrote):

**1) $G$ is the group of conformal symplectic linear maps**

First, some notation. We write $\{e_i\} \cup \{f_j\}$ for the standard basis for $\mathbb{R}^{2n}$, i.e. with $\omega(e_i,e_j) = 0$, $\omega(f_i,f_j) = 0$, and $\omega(e_i,f_j) = \delta_{i,j}$.

Suppose $\phi \in G$. We begin by singling out the isotropic subspaces $A = \mathbb{R}^n \times \{0\}$ and $B = \{0\} \times \mathbb{R}^n$. The following lemma is one way you might think about the group $\mathrm{Sp}(2n,\mathbb{R})$ in the first place: as being defined by its action on an isotropic subspace and a choice of isotropic complement.

**Lemma:** Given $\phi$, there is a unique element $\psi \in \mathrm{Sp}(2n,\mathbb{R})$ such that the following three properties hold:

- $A = (\psi \circ \phi)(A)$, with $(\psi \circ \phi){\big|}_{A} = \mathrm{id} \colon A \rightarrow A$
- $B = (\psi \circ \phi)(B)$

Given this lemma, we have that $\psi \circ \phi = \begin{pmatrix}\mathrm{id} & 0 \\ 0 & T\end{pmatrix}$, and it suffices to find conditions on $T \in \mathrm{GL}(n,\mathbb{R})$ so that isotropic subspaces are preserved by $\psi \circ \phi$. It suffices to consider just two cases:

- For $i \neq j$, the span of $e_i$ and $f_j$ is isotropic, so its image under $(\psi \circ \phi)$, the span of $e_i$ and $Tf_j$, must also be isotropic. This proves that $T$ is diagonal.
- Also for $i \neq j$, we can use the subspace spanned by $e_i + f_j$ and $e_j + f_i$, which will give us the condition that the diagonal entries of $T$ are all equal. Hence $T = c \cdot \mathrm{id}$ for some constant $c \neq 0$.

Finally, we see that with $T = c \cdot \mathrm{id}$, $(\psi \circ \phi)^*\omega = c \cdot \omega$, which clearly preserves isotropic subspaces. So this property completely characterizes $G$.

In terms of nomenclature, $G$ might be called the group of **conformally symplectic** linear operators, and the nonlinear theory would fall under the umbrella of **(locally) conformal symplectic geometry**, which is slightly more general than what you ask (i.e. $\omega$ itself does not need to be symplectic, only (locally) conformally symplectic, which is the more natural setting). I know only a little about the theory, and am far from an expert on the history, so I'll say nothing more that there exists literature on the subject.

As for computations of classical invariants, we see that $G \cong \mathbb{R}^* \times \mathrm{Sp}(2n,\mathbb{R})$, so there's nothing interesting to say about computing its standard invariants that can't already be said for $\mathrm{Sp}(2n,\mathbb{R})$.

**2) $G=H$**

Suppose $\omega(v,w) \neq 0$ and $\omega(v,x) = 0$. Then $\omega(v,w+tx) \neq 0$ for all $t \in \mathbb{R}$, so $v$ and $w+tx$ always span a symplectic subspace. Hence, if $\phi \in H$, since $\phi$ preserves symplectic subspaces, we require $$0 \neq \omega(\phi(v),\phi(w+tx)) = \omega(\phi(v),\phi(w)) + t\omega(\phi(v),\phi(x))$$ for all $t \in \mathbb{R}$. Hence, we must have $\omega(\phi(v),\phi(x)) = 0$ whenever $\omega(v,x) = 0$. In particular, $\omega$ preserves isotropic subspaces, so $H \subseteq G$. Meanwhile, every conformal symplectic linear map certainly preserves symplectic subspaces, so $G \subseteq H$. Hence $G=H$.

**Remark:** Even though the limit of symplectic subspaces may not be symplectic, as in the comments, this is not enough to show $H$ is closed, since if a sequence of elements of $H$ realizes this collapsing of a symplectic subspace onto a non-symplectic one, it can (and must by what I've just written) limit to a singular matrix.