Let $\Delta^k$ be the $k$-simplex, embedded in $\mathbb{R}^{k+1}$ in the usual way so that all edges have length $\sqrt{2}$.. For $k\leq 2$, there are obvious ways to subdivide $\Delta^k$ into $2^k$ congruent copies of the simplex $\frac{1}{2}\Delta^k$, with edges of length $1/\sqrt{2}$. For $k=3$ the theory of scissors congruence shows that this is not possible, essentially because the Dehn invariant has scaling behaviour different from that of the volume. We can instead cut off $4$ copies of $\frac{1}{2}\Delta^3$ adjacent to the vertices, which leaves a regular octahedron, then join each face of the octahedron to the barycentre, giving $8$ more tetrahedra with some edges of length $1/\sqrt{2}$ and others of length $1/2$.

For general $k$, if we just use barycentric subdivision, the longest edges (from the barycentre to one of the original vertices) have length $\sqrt{\frac{k}{k+1}}$, whereas the shortest edges (from the barycentre to the barycentre of a face) have length $\frac{1}{\sqrt{k(k+1)}}$, so the ratio is $k$.

We could instead to edgewise subdivision, so the subdivided complex has vertices $e_{ij}=(e_i+e_j)/2$, and there is an edge from $e_{ij}$ to $e_{kl}$ whenever $i\leq k\leq l\leq j$. In this case all edges have length $1$ or $1/\sqrt{2}$. This is much more regular, but the construction is not invariant under the action of the symmetric group $\Sigma_{k+1}$.

Are there other schemes, preferably invariant under the symmetric group, that divide $\Delta^k$ into $k$-simplices that are reasonably close to being congruent? Or is there some conceptual reason why one should not expect this to be possible?