A necessary condition for a manifold with non-empty manifold boundary embedded in $\mathbb{R}^n$ to have its topological boundary coincide with its manifold boundary is that its manifold interior is an open subset of $\mathbb{R}^n$.
Questions:
Is this also a sufficient condition? (Note that I am assuming a prior that the manifold boundary is non-empty -- this does not hold otherwise.)
What conditions are necessary for a sort of "converse" to this to hold? I.e., given a proper open subset of $\mathbb{R}^n$, which is always a smooth embedded submanifold, when is its topological closure an embedded smooth submanifold with (non-empty) boundary?
Attempt: The second question is not as straightforward as I first thought: consider the open hypercube, which is clearly a proper open subset of $\mathbb{R}^n$ -- however, unless $n=1$, its topological closure is a manifold with corners, but not a manifold with boundary.
Since the open hypercube is homeomorphic (diffeomorphic too, right?) to any open ball, whatever conditions need to be placed on such a proper open subset of $\mathbb{R}^n$ for this "converse" to hold must be "more than topological" (I don't know how to phrase that precisely).
Note: this is not a duplicate of this question, it is a follow-up to it.
Context: This answer established that, if an $m$-dimensional manifold with boundary $M$ can be embedded into a Euclidean space $\mathbb{R}^n$ such that its topological and manifold boundaries coincide, then necessarily one has that $m=n$.
This implies in particular that for $n$-dimensional manifolds with or without boundary which cannot be embedded into $\mathbb{R}^n$, the topological and manifold boundaries never coincide when embedded in any Euclidean space. (If the manifold with boundary has empty manifold boundary then of course the topological and manifold boundaries coincide when it is embedded in itself.)
The interior of any such embedded manifold with boundary is always an embedded manifold (without boundary). We also have that the embedded submanifolds of codimension $0$ of any manifold are exactly the open subsets of that manifold (e.g. Proposition 5.1. p.99 of Lee's Introduction to Smooth Manifolds).
Thus we have a necessary condition for the topological and manifold boundaries to coincide, at least when the embedding is inside of $\mathbb{R}^n$ (although I suspect that this holds for embeddings in an arbitrary $n$-dimensional manifold): the manifold interior of $M$ is an open subset of $\mathbb{R}^n$.
However, this is can not be a sufficient condition: if the manifold boundary is empty then this fails because the topological boundary is non-empty (unless the manifold is all of $\mathbb{R}^n$ itself -- e.g. see this answer). It is unclear to me whether this is a sufficient condition if we assume a priori that the manifold boundary is non-empty.