Say that two polyhedra in $\mathbb{R}^3$ have *isomeasures*
(my terminology) if they have:
the same volume,
the same surface area,
the same sum of all edge lengths,
and the same number of vertices.
The generalization to $\mathbb{R}^d$ requires the identical sums of
the $k$-dimensional measures of all $k$-faces on the exterior of the polytope.

For example, a unit cube has volume $1$, surface area $6$, edge-length sum $12$, and $8$ vertices.

Q. Are there incongruent but isomeasure simplicies in dimensions $d \ge 3$?

There are isomeasure triangles:

Here the blue triangle has base length $2$ and height $1$, while the red triangle has base $\frac{1}{2} \left(-1+\sqrt{2}+\sqrt{6 \sqrt{2}-5}\right)$ and height $\frac{1}{2} \left(-1+\sqrt{6 \sqrt{2}-5}+\sqrt{2 \left(6 \sqrt{2}-5\right)}\right)$, approximately $1.14055$ and $1.75354$ respectively.

Gerhard Paseman pointed out (in a comment to a now deleted question) that there are isomeasure prisms, but I have been unsuccessful in constructing isomeasure tetrahedra.