My first thought on reading this question was, "Take the centroid of the $m$ points. It's obviously equidistant from all of them." Hang on. How am I going to prove that? "Come on! How could it not be?! All the points are indistinguishable... in some sense... and taking the mean to get the centroid doesn't distinguish between them either... in some sense... so..." Wait, what does that even mean?
Here is the result of my digging in my heels and attempting to formalize that initial impulse, rather than doing the reasonable thing of looking for a simple and sensible proof via induction like Henning's and Alex's nice answers.
The $m$ given points $\{v_1, \ldots, v_m\}$ are "indistinguishable" in the sense that any two of them, $v_i$ and $v_j$, can be exchanged by an isometry of $\mathbb R^n$ that leaves the rest of the points unchanged.¹
The centroid does not "distinguish" between points in the sense that any permutation of the given points still has the same centroid.
It turns out that we also need the fact that the centroid is preserved under isometries.²
Now, consider the isometry that swaps $v_i$ and $v_j$. Since the new points are a permutation of the original ones, they have the same centroid, say $v_0$. The isometry maps $v_i$ to $v_j$ and $v_0$ to $v_0$. Since it preserves distances, the distance between $v_i$ and $v_0$ is equal to the distance between $v_j$ and $v_0$. The same argument applies to all pairs of points, and we are done.
¹This isometry is simply reflection across the hyperplane that orthogonally bisects the line segment joining $v_i$ and $v_j$. All other points, being equidistant from $v_i$ and $v_j$, lie on this hyperplane and are unchanged by the reflection.
²That is, an isometry maps centroids to centroids. While this statement is eminently plausible, the only direct proof I can think of is that the centroid may be defined as the point minimizing the sum of squared distances to the $m$ given points.