The answer depends on the precise phrasing of the question. Since you used the word "labeled" in the title, I assume, you are asking for a proof of the following:
Proposition. Let $(x_0,...,x_n)$ and $(y_0,...,y_n)$ be $n$-tuples of points in the Euclidean space $E^3$ such that for every pair of indices $i, j\in \{0,...,n\}$, we have
$$
d(x_i, x_j)= d(y_i, y_j)
$$
where $d$ denotes the distance function on $E^3$. Then there exists an isometry $g$ of $E^3$ such that $g(x_i)=y_i, i=0,...,n$. (The same result holds in all the dimensions.)
Proof. I will be identifying $E^3$ with the 3-dimensional real vector space $V$ equipped with an inner product $\langle \cdot, \cdot \rangle$. Pick points $x_0, y_0$ and, via translations in $E^3$, move these to the origin. Thus, from now on, I will be assuming that $x_0=y_0={\mathbf 0}$.
Let $||\cdot ||$ denote the norm on $V$ defined via the above inner product; then $d(x,y)=||x-y||$.
Now, I assume you know from linear algebra that
$$
||p-q||^2= ||p||^2 - 2\langle p, q\rangle + ||q||^2.
$$
Thus, the pairwise distances between the points $x_i$ and $y_j$ determine the inner products
$$
\langle x_i, x_j\rangle= \langle y_i, y_j\rangle,
$$
$1\le i, j\le n$.
I will consider the generic case, when the vectors $x_1,...,x_n$ span the vector space $V$ and leave you to work out the proof in the case when they do not. This means that three of the vectors, say, $x_1, x_2, x_3$ form a basis in $V$. Equivalently, the Gram matrix $(\langle x_i, x_j\rangle)_{1\le i,j\le 3}$ is nonsingular. Thus, the same holds for the vectors $y_1, y_2, y_3$.
Let $A$ denote the unique linear transformation of $V$ such that $Ax_i=y_i, i=1, 2, 3$. Since the Gram matrix $(\langle x_i, x_j\rangle)_{1\le i,j\le 3}$ equals the Gram matrix $(\langle y_i, y_j\rangle)_{1\le i,j\le 3}$, the transformation $A$ is orthogonal, i.e. a Euclidean isometry. Let's prove that for each $k\ge 4$,
$$
Ax_k=y_k.
$$
Indeed, take any vector $x\in V$,
$$
x= t_1 x_1 + t_2 x_2 + t_3 x_3,
$$
for some $t_1, t_2, t_3\in {\mathbb R}$. Then, because $x_1, x_2, x_3$ form a basis in $V$, the scalars $t_1, t_2, t_3$ are uniquely determined by the three inner products
$$
\langle x_i, x\rangle, i=1, 2, 3.
$$
(You just solve a system of linear equations with the unknowns $t_1, t_2, t_3$ with the matrix of the linear system equal to the above Gram matrix.) The same, of course, holds for the basis vectors $y_1, y_2, y_3$. Since
$$
\langle x_i, x_k\rangle=\langle y_i, y_k\rangle, i=1, 2, 3,
$$
it follows that $Ax_k=y_k$, $k=4,...,n$. qed
Lastly, see examples here which show that can go wrong if you consider unlabeled collections of points and distances.
Best Answer
Supposing that you have the complete matrix of mutual distances between any couple of $n$ points in 3D.
Supposing also that the distances have been measured on $n$ physical points, i.e. we are sure that they represent an actual euclidean geometric layout.
If not, we shall check at least that any three of them respect the triangle inequality. More precisely we can check that for any four of them the Cayley-Menger determinant is positive (the volume of the tetrahedron is real).
That premised, the problem falls in the field of trilateration-multilateration so common in surveying, air navigation, radio astronomy, satellite positioning, etc.
Most probably an algorithm has been developed in one of those disciplines to deal with the problem you propose.
However the underlying concept is simple:
- choose three points, verifying that they are not aligned, anf fix the position of the relevant triangle (e.g. on the $x-y$ plane);
- choose a fourth point, verifying it is not coplanar with the previous, and fix its position (e.g. sign of $z$) as the vertex of a base tetrahedron;
- the tetrahedron will allow to univocally fix the position of a fifth point;
- proceed with the other points using the base terahedron or any of the newly fixed.