Suppose I have a topological $n$-manifold $M$ and a homeomorphisms $\varphi : U \to \mathbb{B}^n$ where $\mathbb{B}^n$ is the open unit-ball in $\mathbb{R}^n$ and $U \subseteq M$ is some open set.. Suppose that I have a homeomorphism $\xi : D^n \to D^n$ such that $\xi$ is fixed on the boundary of $D^n$. Then note that $\psi = \varphi^{-1} \circ \xi|_{\mathbb{B}^n} \circ \varphi : U \to U$ is a homeomorphism. Is it then possible to extend $\psi$ to a homeomorphism $\widetilde{\psi} : M \to M$? If not, are there any other conditions we need on $\varphi$ to ensure this is the case?
Similarly when is it possible to extend a homeomorphism $\zeta : \operatorname{Int}(M) \to \operatorname{Int}(M)$ to all of $M$?
Assume $U$ is tamely embedded with boundary $\mathbb{S}^{n-1}$. Here $M$ is assumed to have no boundary. If you let $\xi$ be the identity then we are just talking about when $\phi$ extends to an orientation-preserving homeomorphism on the boundary. If there is such an extension, then you can take a collar neighborhood $N$ around $\mathbb{S}^{n-1}$ and apply the annulus theorem + Schoenflies to construct a homeomorphism on all of $M$ by letting $f \equiv \text{id}_M$ on $N^c$ and homotoping to the identity in the annulus $N$ (which is possible via the degree theory of sphere mappings). Note that homotopic sphere homeomorphisms are homotopic via homeomorphisms.
EDIT: Actually, there's a bit more to it; I forgot about the degree $-1$ case. If you have your homeomorphic extension to the boundary of $U$ but it's the wrong orientation, there may still be a homeomorphism on $M$. For example, reflecting the open disc in the plane, viewed as a subset of the sphere; the whole sphere can be flipped. Assuming $M$ is connected, then any two open balls with tame boundaries can be carried onto one-another via a homeomorphism of $M$, so you want to have the boundary extension + orientation-preserving OR homeo$(M)$ contains an element which flips the orientation on a tame $(n-1)$-sphere which bounds an open ball.
But in general this is not always possible, even in the plane: For example, take a homeomorphism of $\mathbb{D}^2$ to itself that preserves the distance of any point to zero, but which rotates faster and faster unboundedly as you approach $\mathbb{S}^1$. Then as you approach the unit circle via a straight line from the center its trajectory will be dense in $\mathbb{S}^1$ so won't have an extension. You can do similar things in any dimension larger than $1$; in the even dimensions use similar rotations, and in odd dimensions you can "kneed the dough back and forth" more and more as you approach the boundary - maybe someone can give an explicit map in the odd dimensions which is easier to write down than what I'm talking about.
If your sphere isn't tamely embedded then even having the extension to the boundary won't be strong enough to imply what you want, unfortunately. Extension of homeomorphisms isn't a 'solved problem' in topological manifolds of dimension higher than $2$ - there won't be a simple answer, only partial and technical results which may only hold in specific dimensions.
If the boundary of $U$ self-intersects in $M$ but is 'otherwise tame' in some nebulous sense then I have no idea. There would be all sorts of cases depending on how it intersected with itself, probably.