When is an $n$-dimensional manifold characterized by its $m$-dimensional submanifolds? For which $m$, $n$ (if any) is the following true: if $M$ and $M'$ are smooth manifolds of dimension $n$, and $\Phi$ is a bijection from $M$ to $M'$ such that for any subset $S$ of $M$, $\Phi(S)$ is an embedded submanifold of $M'$ of dimension $m$ iff $S$ is an embedded submanifold of $M$ of dimension $m$, then $\Phi$ is a diffeomorphism?
This is clearly false when $m=0$ and when $m=n$.  I thought it looked plausible when, for example, $m=1$ and $n\geq 2$; but I can't see how to prove it.  
 A: The claim is false as stated, but I think a refinement can fix the problem.  (Not sure what the correct refinement is yet.)
My counterexample is essentially a generalization of taking $\Phi: \mathbb R \to \mathbb R, x \mapsto x^3$.  For $n=2, m=1$, we will take $M=M' = \mathbb R^2 = \mathbb C$.  Using complex coordinates $z = x+iy$, let $\Phi(z) = z |z|^2$.  This is a diffeomorphism of $\mathbb C \setminus \{ 0 \}$, and is a homeomorphism of $\mathbb C$.  The inverse map is $\Phi^{-1}(z) = z |z|^{-2/3}$ (with $\Phi^{-1}(0) = 0$).
I will show that if $S$ is embedded in $\mathbb C$ then $\Phi(S)$ is embedded.  The other direction follows the same argument, but using the inverse function instead.
If $S$ avoids $0$, there is nothing to check.  If $S$ intersects $0$, we reduce to the local problem of a smooth curve $\gamma: [-1,1] \to \mathbb{C}$, $\gamma(0) = 0$, $\gamma'(t) \ne 0$.  We then check that $\gamma( t^{1/3})$ satisfies that $\Phi( \gamma( t^{1/3}))$ is smooth.  (The easy way to see this is to note that $\gamma(t) = ut + O(t^2)$ for a non-zero constant $u$, and thus $\Phi( \gamma(t) ) = u|u|^2 t^3 + O(t^4).)
A: (This is a bit long for a comment, but it's not a full answer)
Here's where I would start when looking at this question.

Theorem: The category of smooth manifolds is a full subcategory of the category of Frölicher spaces.

First, we need to know what a Frölicher space is!  It is triple $(X,C,F)$ where $X$ is a set, $C \subseteq \operatorname{Map}(\mathbb{R},X)$ is a family of curves in $X$ and $F \subseteq \operatorname{Map}(X,\mathbb{R})$ is a family of functionals on $X$.  Note that these are just set maps.  The curves and functionals have to satisfy a compatibility condition:


*

*$\alpha \in C$ if and only if $\phi \circ \alpha \in C^\infty(\mathbb{R},\mathbb{R})$ for all $\phi \in F$, and

*$\phi \in F$ if and only if $\phi \circ \alpha \in C^\infty(\mathbb{R},\mathbb{R})$ for all $\alpha in C$.


A morphism of Frölicher spaces is a set map $f \colon X \to Y$ which satisfies the following (equivalent) conditions:


*

*$f \circ \alpha \in C_Y$ for all $\alpha \in C_X$,

*$\phi \circ f \in F_X$ for all $\phi \in F_Y$,

*$\phi \circ f \circ \alpha \in C^\infty(\mathbb{R},\mathbb{R})$ for all $\alpha \in C_X$ and $\phi \in F_Y$.


The chain rule is enough to show that there is a functor from the category of smooth manifolds to that of Frölicher spaces, where we assign to a smooth manifold $M$ the Frölicher space $(M, C^\infty(\mathbb{R},M), C^\infty(M,\mathbb{R}))$.  We want to show that this embeds the category of smooth manifolds as a full subcategory.
Firstly, that it is faithful is obvious since both categories are concrete (that is, equipped with a faithful functor to $\operatorname{Set}$) and the functor preserves this structure.
So we just need to show that it is full.  As both categories are concrete (and the functor preserves this structure), it is sufficient to show that if $f \colon M \to N$ is a set map which is not smooth, then it is not a morphism in the Frölicher category.  So let $M$ and $N$ be two smooth manifolds and $f \colon M \to N$ a set map which is not smooth.  Then there is a chart for $M$ and a chart for $N$ which detects this non-smoothness.  That is to say, there are smooth maps $\theta_M \colon U_M \to V_M \subseteq M$ and $\theta_N \colon U_N \to V_N \subseteq N$ such that $g = \theta_N \circ f \circ \theta_M$ both makes sense and is not smooth.
Now $g$ is a map from an open subset of some Euclidean space to an open subset of some Euclidean space.  A map in to an open subset of a Euclidean space is smooth if and only if it is smooth when considered as a map into the ambient space, and then a map into a Euclidean space is smooth if and only if all the compositions with the projections are smooth.  Thus we can find some projection $\pi_i \colon \mathbb{R}^n \to \mathbb{R}$ such that $\pi_i \circ \iota \circ g \colon U_M \to \mathbb{R}$ is not smooth.  Furthermore, as $U_M$ is an open subset of some Euclidean space, we can find an open disc such that the restriction of this map to that disc is not smooth, and then composing with a diffeomorphism of the ambient space to that disc provides us with a non-smooth map $\mathbb{R}^m \to \mathbb{R}$.
Now comes the magic step.  We use a result of Jan Boman which says that a map $\mathbb{R}^m \to \mathbb{R}$ is smooth if and only if its compositions with all smooth curves are smooth.  That is, $h \colon \mathbb{R}^m \to \mathbb{R}$ is smooth if and only if $h \circ \gamma \colon \mathbb{R} \to \mathbb{R}$ is smooth for all $\gamma \in C^\infty(\mathbb{R},\mathbb{R}^m)$.  So as our map was not smooth, there is a smooth curve $\gamma \colon \mathbb{R} \to \mathbb{R}^m$ which detects this non-smoothness.  We can now unpick all the constructions to transfer this curve to a smooth curve in $M$, similarly we can transfer our projection to a smooth functional on $N$, and produce a smooth curve $\alpha \in C_M$ and functional $\phi \in F_N$ such that $\phi \circ f \circ \alpha$ is not smooth.  Hence $f$ is not a morphism in the category of Frölicher spaces.
This proves the theorem.

Now the question as stated asks about embedded submanifolds, not all maps from, say, $\mathbb{R}$.  So the place to look is to see if Boman's result works if one is allowed to only test with curves that are parametrisations of submanifolds.  As this is a local result, and immersions are local embeddings, it is enough to test with curves $\alpha \in C^\infty(\mathbb{R},M)$ with non-vanishing first derivative.
My instinct would be to say that the result still holds, but I don't have a good enough grasp of the details of Boman's proof to be sure.  I do know that you don't need all smooth curves for it to work.  For example, you can leave out all curves that have non-negligible intersection with, say, the $x$-axis.
References


*

*Boman, J. (1967). Differentiability of a function and of its compositions with functions of one variable. Math. Scand., 20, 249–268.

*Kriegl, A. and M., Peter W. (1997). The convenient setting of global analysis (Vol. 53). Mathematical Surveys and Monographs. Providence, RI: American Mathematical Society.

*nLab pages on, starting at Frölicher spaces.
