Reference Request: Regge Symmetry "Angle-Edge" Duality
Solution 1:
Responding to my own question to get it out of the "Unanswered" queue (after eighteen months there!) ...
Email correspondence with a mathematician in the know confirms that the result of the prescribed transformation of dihedral angles is equivalent to the result of applying the transformation to corresponding edge lengths. Also confirmed: verification invariably seems to involve a long, hard symbolic slog.
Some notes ...
Let $(a_\star, b_\star, c_\star, a^\prime_\star, b^\prime_\star, c^\prime_\star)$ be the edge lengths of a tetrahedron after a transformation of the dihedral angles. It's not too terribly difficult to verify these relations: $$\begin{align} \overline{b_\star}\overline{b^\prime_\star} + \overline{c_\star}\overline{c^\prime_\star} &= \overline{b}\overline{b^\prime} + \overline{c}\overline{c^\prime} \qquad\qquad \overline{b_\star}\overline{b^\prime_\star} - \overline{c_\star}\overline{c^\prime_\star} = -\ddot{b}\ddot{b^\prime} + \ddot{c}\ddot{c^\prime} \\ \ddot{b_\star}\ddot{b^\prime_\star} + \ddot{c_\star}\ddot{c^\prime_\star} &= \ddot{b}\ddot{b^\prime} + \ddot{c}\ddot{c^\prime} \qquad \qquad \ddot{b_\star}\ddot{b^\prime_\star} - \ddot{c_\star}\ddot{c^\prime_\star} = -\overline{b}\overline{b^\prime} + \overline{c}\overline{c^\prime} \\ \ddot{b_\star}\ddot{c^\prime_\star} + \ddot{c_\star}\ddot{b^\prime_\star} &= \ddot{b}\ddot{c^\prime} + \ddot{c}\ddot{b^\prime} \qquad\qquad \ddot{b_\star}\ddot{c_\star} + \ddot{b^\prime_\star}\ddot{c^\prime_\star} = \phantom{-}\ddot{b}\ddot{c} + \ddot{b^\prime}\ddot{c^\prime} \end{align}$$
where $\overline{x} := \sinh{x}$ and $\ddot{x} := \cosh{x}$.
The first two rows are symmetric in the edge pairs $\{b_\star,b^\prime_\star\}$ and $\{c_\star, c^\prime_\star\}$ and cannot by themselves distinguish between those values. At best, they imply $$ \{2 b_\star, 2 b^\prime_\star\} = \pm \left( - b + b^\prime \right) + c + c^\prime \qquad \{2 c_\star, 2 c^\prime_\star\} = b + b^\prime \pm \left(- c + c^\prime \right) $$ The third row breaks some symbolic symmetry, and we can deduce that $b_\star$'s "$\pm$" matches that of $c_\star$ (and that $b^\prime_\star$'s matches $c^\prime_\star$'s).
To resolve the final sign ambiguity, one can verify that $b+b_\star = c+c_\star \;(= b^\prime + b^\prime_\star = c^\prime+ c^\prime_\star)$. One computationally-arduous strategy is to show that $\tanh(b+b_\star) = \tanh(c+c_\star)$. (For me, targeting $\tanh$ turned out to be less difficult than considering $\sinh$ or $\cosh$.)