I asked this same question on Math Overflow but I might get a quicker response if I post it here as well.
Here is a preamble. Suppose we have a symplectic manifold $(M,d\lambda)$ with an exact symplectic form. By Stokes theorem, $M$ must have nonempty boundary. An exact symplectomorphism $\phi:M \to M$ is a diffeomorphism such that $\phi^* \lambda - \lambda = df$; i.e. the difference is an exact 1-form.
Now, in some notes by Seidel, he considers the monodromy map $\phi$ one obtains for the Milnor fiber $F=\{x^2 +y^2 = \delta; |x|^2+|y|^2 \leq \epsilon \} \subset \mathbb{C}^2$ for $|\delta|$ much smaller than $\epsilon$. This $F \cong S^1 \times [0,1]$. Letting $k \in \mathbb{Z}^+$, one can isotope $\phi^k$ so that it has $k-1$ fixed circles in $S^1 \times [0,1]$. In fact, if we're using a smooth isotopy, we can get rid of all the fixed points. However, Seidel says that if our isotopy must pass through exact symplectomorphisms, then the best we can do is to disintegrate each fixed circle into a pair of fixed points.
This reminds me of the fact that on the cylinder $T^* S^1$, the zero section $L$ cannot be displaced by Hamiltonian isotopy $\psi$. Indeed, the image of $L$ will intersect $L$ in at least two points because the signed area of the 2-chain bounded by $L$ and $\psi(L)$ must be zero. If instead, we move $L$ by an exact symplectomorphism, we can confirm Seidel's statement that there will be at least two fixed points by again, seeing that the signed area of the 2-chain must be zero.
So here are some questions.
Question 1: What is the relationship between exact symplectomorphisms and Hamiltonian diffeomorphisms on $(M,d\lambda)$? It seems they have a close relationship by the following argument:
Suppose that $\phi_t$ is an isotopy of exact symplectomorphisms. So $\phi_t^* \lambda = \lambda + df_t$. Differentiating both sides with respect to $t$ and then setting $t=0$, we get $\mathcal{L}_{X_t} \lambda = d i_{X_t} \lambda + i_{X_t} d\lambda = d\dot{f}_0$. Here, $X_t$ are vector fields defined by $\dot{\phi}_t = X_t \circ \phi_t$.
Hence, $i_{X_t} d\lambda = d(\dot{f}_0 - i_{X_t} \lambda)$, an exact 1-form. This means that $X_t$ is a Hamiltonian vector field. If I didn't make any mistakes, I think this tells us that the isotopy $\phi_t = \phi_0 \circ \psi_t$ where $\psi_t$ is a Hamiltonian isotopy. In particular, it appears that on exact symplectic manifolds, the identity component of the group of exact symplectomorphisms is contained in the group of Hamiltonian diffeomorphisms.
Question 2: It seems that many authors require that the $f$ in $\phi^* \lambda = \lambda + df$ has compact support away from the boundary of $M$. Why is this often assumed? Is it because if we don't assume this, then the exact symplectomorphism might not have any fixed points? The argument above suggests that at least when the exact symplectomorphism is in the identity component, it cannot displace Lagrangian submanifolds (in appropriate settings where Arnold's conjecture for Lagrangian intersections holds) and so we would have fixed points.
Question 3: It seems an answer to question 2 may answer this but why is it often assumed that the polynomial used to define the Milnor fiber has an isolated singularity? If it does not, then one can get monodromy that is not compactly supported away from the boundary of the fiber so we would have the same issues as in question 2. Are there further issues?