Let me first try to make sense of your question. First of all, the notion of a differential form is meaningless for general metric spaces $(X,d)$. However, one can still talk about Borel measures $\mu$ on the topological space $X$ (topologized using the metric $d$). The Borel condition still leaves too much freedom since we did not use the metric $d$ (only the topology). The most common condition these days is the one of a metric measure space which ties nicely $d$ and $\mu$ and allows one to do quite a bit of analysis on $(X,d,\mu)$ similarly to the analysis on the Euclidean $n$-space $E^n$. (The literature on this subject is quite substantial, just google "metric measure space".)
Definition. A triple $(X,d,\mu)$ (where $d$ is a metric on $X$ and $\mu$ is a Borel measure on $X$) is called a metric measure space if the measure $\mu$ is doubling with respect to $d$, i.e. there exists a constant $D<\infty$ such that for every $a\in X, r>0$, we have
$$
\mu(B(a, 2r))\le D \mu(B(a,r)),
$$
where $B(a,R)$ denotes the closed ball of radius $R$ centered at $a$.
Example. Every closed subset of $E^n$ (equipped with the restriction of the Euclidean metric) is a complete doubling metric space.
The basic existence result for doubling measures $\mu$, proven in
"Every complete doubling metric space
carries a doubling measure", by J. Luukainen, E. Saksman, Proc. Amer. Math. Soc. 126 (1998) p. 531–534, is the following:
Theorem. A complete metric space $(X,d)$ carries a doubling measure $\mu$ if and only if the metric $d$ is doubling, i.e. there exists a constant $C$ such that every ball of radius $2r$ in $X$ is covered by at most $C$ balls of radius $r$.
The proof of this theorem is not long but by no means trivial.
You can say an immense amount.
This is the simplest possible case of the groups $\text{Symp}(M)$, where $M$ is a symplectic manifold. First allow me to say some generalities on those.
If $M$ is a connected symplectic manifold, the group $\text{Symp}(M)$ acts transitively on ordered sequences $(x_1, \cdots, x_n) \in M^n$ so that all $x_i \neq x_j$ (unless $i = j$). This is called "$n$-transitivity", and should be read as saying "I can take any $n$ distinct points to any other $n$ distinct points". This is easily implied by the claim that the group $\text{Symp}_c(M)$ of compactly supported symplectomorphisms (those that are the identity outside of a compact set) acts transitively on $M$. I will leave this implication as an exercise to you, but explain why $\text{Symp}_c$ acts transitively.
This is best discussed by the construction of Hamiltonian vector fields. Given a function $f$ on a domain, one defines $X_f$ to be the unique vector field with $\omega(X_f, Y) = df(Y)$ for all vector fields $Y$. Then flow along $X_f$ preserves the symplectic form, and $\text{supp}(X_f) \subset \text{supp}(f)$. Now it suffices to show that the orbits of $\text{Symp}_c$ are open sets, as then we would have a decomposition of $M$ into disjoint open sets; thus orbits would be all of $M$. So we only need to construct a Hamiltonian vector field taking us to a nearby point.
So pick a standard symplectic chart around each $x \in M$; we will show that in $U \subset \Bbb R^{2m}$, there is a Hamiltonian vector field, compactly supported in $U$, flowing $0$ to any point sufficiently close to $0$.
To do this, if $v \in \Bbb R^{2m}$ is any vector, let $f_v(x) = \langle v, x\rangle$; because this is linear, we also have $df_v = \langle v, -\rangle$ at every point. Because the symplectic form here is $\omega(X_v, x) = \langle X_v, Jx\rangle = \langle -JX_v, x\rangle$, and by definition $\omega(X_v, x) = df_v(x) = \langle v, x\rangle$, we find that we must have $X_v = Jv$, where $J$ is the multiplication-by-$i$ map on $\Bbb R^{2m} = \Bbb C^m$.
In particular, $X_v$ is just "flow in the constant direction $Jv$". Replacing $f_v$ with $\beta f_v$, where $\beta$ is a bump function equal to $1$ near $0$ and supported inside a compact subset of $U$, we obtain a corresponding vector field $X'_v$ which is compactly supported inside $U$, but for small $t$ the time-$t$ flow just sends $0$ to $tJv$, and hence an orbit of $\text{Symp}_c(M)$ is open, as desired.
In particular, note that $\text{Symp}_c(M)$ must be infinite-dimensional, given that it has a surjective map to an open subset of $M^n$ for all $n$ (that is, $n$-transitivity means this group must be gigantic).
In dimension $n = 1$ (real dimension 2), however, things become much more simple: $\text{Symp}(M) = \text{Vol}(M)$, the group of diffeomorphisms fixing a particular volume form. This group similarly acts $n$-transitively in any dimension, and is in many ways much simpler than the symplectomorphism group. In fact, the original Moser trick proves a few facts: 1) any two volume forms $\omega_1, \omega_2$ of the same volume are related by some diffeomorphism $f$; and 2) the full group $\text{Diff}(M)$ is homotopy equivalent to the subgroup $\text{Vol}(M)$ of diffeomorphisms preserving a fixed volume form. So there is "no new topology to be had" in studying volume-preserving diffeomorphisms. This is very much not true in the symplectic case (in higher dimensions, of course).
None of these are related to analyticity (real or complex).
For lots of interesting facts about many related groups (Diff, Symp, Vol, ...) see Banyaga's book on the subject. One famous fact: when $M$ is compact, the connected component of the identity $\text{Vol}_0(M)$ is a simple group. (In the symplectic case, one has to further restrict to the kernel of a certain map called the flux homomorphism; see Banyaga's book or the shorter article by McDuff.)
Best Answer
There are certainly examples of symplectomorphisms which do not preserve $J$ or $g$. In fact, for any closed symplectic manifold the symplectomorphism group is infinite dimensional. The group of complex automorphisms and isometries is finite dimensional.
Here is a sketch of the proof that the symplectomorphism group is infinite dimensional. Consider a smooth function $H : M \rightarrow \mathbb{R}$. Then, since symplectic forms are non-degenerate, one may check that there is a unique vector field $X_{H}$, satisfying $\omega(X_{H},\cdot) = dH$, called the Hamiltonian vector field ascosiated to H. The flow of the vector field is by symplectomorphisms, by Cartan's magic formula, which I leave for you to verify. Since the space of smooth functions $C^{\infty}(M,\mathbb{R})$ is infinite dimensional, this argument along with a few additional technicalities shows that the symplectomorphism group is infinite dimensional. See "Introduction to symplectic topology" by McDuff and Salamon for a formal proof.