Let $M$ be a closed and simply-connected 4-manifold and let $f: M^4 \to M^4$ be a diffeomorphism such that $f^*: H^*(M;\mathbb{Z})\to H^*(M;\mathbb{Z})$ is the identity map. Moreover, let $\Sigma \subset M$ be an oriented closed surface such that $f|_\Sigma = \mathrm{Id}$. Does that imply that up to isotopy, $f$ is the identity on a neighborhood of $\Sigma$, i.e. there exists a diffeomorphism $f'$ such that $f$ is smoothy isotopic to $f'$ and $f'|_{N(\Sigma)} = \mathrm{Id}$?
Now one can use some basic obstruction theory to conclude that the above diffeomoprhism $f$ will induce an obstruction class $\phi : \Sigma \to \operatorname{Diff}^+(D^2) \sim S^1$. And whenever this obstruction class vanishes i.e. $\phi$ is nullhomotopic, the above diffeomorphism can be made the identity on a neighborhood of $\Sigma$. For example, if $\Sigma= S^2$. But this map $\phi$ is not necessarily nullhomotopic for surfaces with higher genus. However, I believe that still the existence of a non-trivial such map $\phi$ is impossible, and somehow it has to do something with the fact that $f^*= \mathrm{Id}$.
EDIT- Note that it is possible to construct a diffeomorphism that induced the identity map on a submanifold but one cannot make it identity on a neighborhood by a smooth isotopy. For example, take a solid torus $S^1\times D^2$ and consider the Dehn twist along the $\partial D^2$. This slef-diffeomorphism fixes the core $S^1\times \{0\}$. However one cannot make this map smoothly isotopic to identity on the neighborhood, otherwise, Dehn twist on $T^2$ will be (pseudo)isotopic to the identity, which is not possible. The issue here is that the above mentioned obstruction class $\phi$ corresponds to the submanifold $S^1\times \{0\}$ is actually not null-homotopic.
