Let $x_i,x'_i \in \mathbb{R}^n$, $i=1,...,k $, $k \le n$ and suppose that
$$ d_{ij}' =\| x_i'-x_j'\|\le \| x_i-x_j\|=d_{ij} \tag{1}$$
$$ \| x_i\|= \| x_i'\|\tag{2}$$
It is known that there exist a continuous family of $k$-tuples $x_i^t \in \mathbb{R}^n$ such that $x_i^0=x_i,x_i^1=x_i'$ and $d_{ij}^t=\| x_i^t-x_j^t\|$ is decreasing in $t$.
A proof is sketched here (In lemma 2.A. I give some more details below. In fact the process described by Gromov only recovers the given configurations up to an isometry of $\mathbb{R}^n$, as far as I can see).
Question: Is there such a continuous deformation for $n+1$ points in $\mathbb{R}^n$?
(It seems known that there is no a continuous realization of every deformation for $n+2$ points in $\mathbb{R}^n$, for example for $n=2$-i.e $4$ points in the plane).
Details for the $k \le n$ proof:
Define $A_{ij}^t= (1-t)\langle x_i,x_j \rangle+ t\langle x_i',x_j' \rangle$.
$A^t$ is positive semidefinite $k \times k$ matrix. Let $\sqrt{A^t}$ be its unique positive semidefinite square root . Define $x_i^t \in \mathbb{R}^n$ to be the $i$-th column of $\sqrt{A^t}$, after appending $n-k$ zeros.
Then, by construction, we get $A_{ij}^t=\langle x_i^t,x_j^t \rangle$.
Since $\langle x_i^t,x_i^t \rangle=A_{ii}^t=\| x_i\|^2= \| x_i'\|^2$,
$$ (d_{ij}^t)^2 = \langle x_i^t,x_i^t \rangle+\langle x_j^t,x_j^t \rangle-2 \langle x_i^t,x_j^t \rangle=\| x_i\|^2+\| x_j\|^2-2 \langle x_i^t,x_j^t \rangle.$$
Combining this with the fact that $$ \langle x_i,x_j \rangle=\frac{1}{2}(\| x_i\|^2+\| x_j\|^2-\| x_i-x_j\|^2) \le \langle x_i',x_j' \rangle$$ (so $A_{ij}^t$ is increasing in $t$), we get that $d_{ij}^t$ is decreasing as required.
In this way we do not always get the original points at the endpoints $t=0,1$ but we do recover them up to an isometry of $\mathbb{R}^n$ which is enough.
(It is interesting if there is a deformation process preserving the original initial and final configurations. We can assume W.L.O.G that $x_i,x_i'$ live inside a copy of $\mathbb{R}^k$ embedded in $\mathbb{R}^k$ (say the last $n-k$ coordinates are zeros, but I am not sure this is enough).
(1) In $\mathbb{R}^2$ consider vertex points $a_i=a_i'$ of regular triangle $T$ of side length 1 and a point $a_4$ outside of $T$. Here $a_4$ has a large distance from $T$. And $a_4'$ is center of $T$.
So under deformation, at some time, $a_4$ must be in $B_\frac{1}{2}(a_i)$ for some $i$ Since $d(a_4',a_i) > \frac{1}{2}$, there is no continuous deformation.
More explanation : If $d_{ij}(t)$ is distance between $a_i(t),\ a_j(t)$ where $a_i(0)=a_i,\ a_i(1)=a_i'$, then $d_{ij}(t),\ 1\leq i<j\leq 3$ is constant.
For $1\leq i\leq 3$, $$ \frac{1}{\sqrt{3}} \leq d_{i4}(t)$$ is decreasing
For all $1\leq i\leq 3$, $a_4(1)$ with $d_{i4}(1)=\frac{1}{\sqrt{3}}$ is unique. That is $a_4$ must go into $T$ Hence $a_4(t)$ is in $[a_i(t)a_j(t)]$ for some $t$ and some $i,\ j$. So $\min\ \{ d_{i4}(t),d_{j4}(t) \}\leq \frac{1}{2}$
(2) And on $S^2$, there is no deformation for $\{p_i\},\ \{p_i'\}$ whose positions are similar to the above.