Is a complete set of labelled pairwise distances between all points sufficient to specify a configuration in 3d space?

This question is motivated by the recent AlphaFold protein structure prediction model by Google DeepMind. In order to predict the 3d structure of a protein molecule the DeepMind team first estimate a matrix of pairwise distances between each amino-acid residue on the protein molecule. This made me wonder if the set of pairwise distances between points is sufficient to fully describe the positions of the points relative to each other. Clearly, any configuration of points - with each point represented by vector (x,y,z) - can be transformed by rotation, translation and reflection and still retain the same pairwise distances, but I want to know if it is possible that a set of pairwise distances could allow for more than one configuration of points such that the configurations cannot be made the same by applying any combination of reflection, translation or rotation? My question only applies to 3-dimensional space. This process is sometimes referred to as multidimensional scaling when used as a dimensionality reduction technique, but the theory I read on this subject does not indicate whether the resultant configuration is unique for a given set of pairwise distances.

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.