Suppose we have a 1D smooth manifold that locates on the x-y plane, and we have some information about it. Can we identify the manifold uniquely? Suppose we have a 1D smooth manifold that locates on the x-y plane, and we know the number of intersections for any straight line on the x-y plane with the manifold. Can we recover its shape uniquely?

 A: First of all, one has to clarify what the "number of intersections" means: Even for (smooth) simple closed curves in the plane, the intersection of a line with the curve can be infinite: Both countably infinite and of cardinality of continuum. A more precise version of your question then is:

Suppose that $A, B$ are 1-dimensional smooth submanifolds of ${\mathbb R}^2$ such that for every affine line $L\subset  {\mathbb R}^2$ the sets  $A\cap L$ and $B\cap L$ have the same cardinality. Does it follow that $A=B$?

The answer to the question posed in this form is negative if we allow noncompact submanifolds. The simplest examples are given by Archimedean spirals, which, in polar coordinates $(r,\theta), r>0, \theta\in (0,\infty)$, are written as
$$
r= b\theta, b> 0. 
$$
Each spiral has countably infinite intersection with every affine line in the plane, but, (for instance) choosing different values of the parameter $b$, one obtains different spirals (which are not even congruent to each other by a Euclidean rigid motion).
However, one can prove that if $A$ and $B$ are compact smooth 1-dimensional submanifolds, then your question indeed has positive answer. Let me know if you are interested in the proof.

Edit. Here is how to handle simple closed curves.
Definition. Given a smooth curve $C$ in the plane and a (straight) line $L$ in the plane, define $n_C(L)$ to be the cardinality of the intersection $C\cap L$.
In general, this cardinality is infinite (either countably-infinite or cardinality of continuum). In the special case, when $C$ is a simple closed real-analytic curve, the function $n_C$ takes only finite values.   I will need definitions.
Definition. A line $L$ is said to meet $C$ transversally at a point $p\in C\cap L$, if the line $L$ is not tangent to $C$ at $p$.
A line $L$ is said to be transversal to $C$ if it meets $C$ transversally at every point $p\in C\cap L$ (this intersection could be empty).
From now on, I will only consider only smooth simple closed curves $C$ in the plane.
Important properties of lines meeting $C$ transversally are:

*

*$n_C(L)<\infty$.


*If $L_1$ is another line sufficiently close to $L$ (one can state this closeness for instance in terms of coefficients of linear equations defining lines), then $L_1$ also meets $C$ transversally and $n_C(L)=n_C(L_1)$.
Definition. Suppose that $L$ is a line tangent to a smooth simple curve $C$ at a point $p$. Then the intersection of $L$ and $C$ at $p$ is called nondegenerate or the first order of contact if the following holds. Choose Cartesian $xy$- coordinates in the plane so that $L$ is the $x$-axis and $p=(0,0)$. Then a small arc of $C$ containing $p$ is the graph of a smooth function $y=f(x)$. The 1-st order of contact means that $f'(0)=0$ (the tangency condition) and $f''(0)\ne 0$.
In particular, for first order of contact at $p$, a small arc of $C$ containing $p$ lies entirely on one side of $L$ (I'll call this the $C$-side).
Here is an important property of nondegenerate intersections:
Lemma 1. Suppose that $L$ is a line which intersects $C$ at points $p_1,...,p_n$ (thus, $n_C(L)=n$), such that the intersections are transversal at $p_1,...,p_{n-1}$ and the intersection at $p_n$ is the 1st order of contact. Then, for every line $L_1\ne L$ parallel to $L$ and sufficiently close to $L$, $n_C(L_1)= n\pm 1$. Moreover, it is $+1$ if and only if $L_1$ lies on the $C$-side of $L$.
Theorem. Suppose that $C_1, C_2$ are distinct (smooth simple closed) planar curves. Then there is a line $L$ such that
$n_{C_1}(L)\ne n_{C_2}(L)$.
Proof.
Lemma 2. Let $A_1, A_2$ be smooth arcs in $R^2$ such that for every point $p\in A_1$ the tangent line $L_p$ to $A_1$ at $p$ is also tangent to $A_2$ at some point. Then either $A_1$ is a straight line segment or $A_1\subset A_2$.
Suppose that $L$ is a line which meets $C_2$ transversally, while it meets $C_1$ as in Lemma 1. Assume that $n_{C_1}(L)=n_{C_2}(L)=n$ (otherwise, we would be done). Then, replacing $L$ with a different line $L_1$, parallel to $L$ and sufficiently close to $L$, we obtain:
$$
n_{C_1}(L_1)= n\pm 1, n_{C_2}(L_1)= n. 
$$
This would imply  $n_{C_1}(L_1)\ne n_{C_2}(L_1)$! It remains to find a line $L$ intersecting $C_1, C_2$ as described above.
Here is a proof of existence of such $L$ in the case of real-analytic curves $C_1, C_2$. (The real-analyticity assumption cuts down on the number of technicalities in the proof, which is already quite long and does not even includes proofs of Lemmata 1 and 2, but is not really essential.)
For each $p\in C_1$, let $L_p$ denote the line tangent to $C_1$ at $p$. Since $C_1$ is real-analytic, there is a finite subset $E_1\subset C_1$, such that for all $p\in C_1\setminus E_1$,  the line $L_p$ has 1st order of contact with $C_1$ at $p$. It is a priori, possible, that for some $p\in C_1\setminus E_1$, the line $L_p$ is tangent to $C_1$ at another point $q\ne p$. Lemma 2 (applied to pairs of arcs $A_1, A_2$ in $C_1$) implies that the subset $E_2\subset C_1$ of such points $p$ is finite (I again use real-analyticity assumption here): Since $p\ne q$ the option $A_1\subset A_2$ is impossible, while no arc of $C_1$ can be a straight-line segment, since $C_1$ is assumed to be real-analytic.
Now, consider points $p\in C_1\setminus (E_1\cup E_2)$. Applying Lemma 2 again (now, to pairs arcs
$A_1\subset C_1, A_2\subset C_2$) and using real-analyticity one more time, we see that there is a finite subset $E_3\subset C_1$ such that for all $p\in C_1\setminus (E_1\cup E_2\cup E_3)$, the line $L_p$ meets $C_2$ transversally. Indeed, the case of an arc in $C_1$ which is a straight-line segment is ruled out as above, while $A_1\subset A_2$ would imply (by real-analyticity of $C_1, C_2$!) that $C_1=C_2$, which we assumed is not the case.
Thus, for every $p\in C_1\setminus (E_1\cup E_2\cup E_3)$, the line $L_p$ meets $C_2$ transversally, while it meets $C_1$ as in Lemma 1. As was noted above, this proves existence of  a line $L$ such that $n_{C_1}(L)\ne n_{C_2}(L)$. qed
