Imagine a single parameter real-valued non-zero function $f$. For any complex number $Z$, let
$Z \rightarrow Z' = \Re(Z) f + i\Im(Z)/f $
Calculate $Z^2$ and $Z'^2$. Because the imaginary portion of the square is the product of the function f and its inverse, it is unchanged:
$\Im(Z^2) = \Im(Z'^2)$
Is there a name for this symmetry?
Your $Z'$ all belong to the same equivalence class as Z, if we define the equivalence relation as $z,z'\in\mathbb{C};\quad z\equiv z' \implies \Im(z^2)=\Im(z'^2)$.
We can prove this by proving $[z,z']$ is reflexive, symmetric and transitive.
Reflexive: $[z,z]\implies [z,z]$
Symmetric: $[z,z']\implies [z',z]$
Transitive: $[x,y] \land [y,z] \implies [x,z]$
all of which have trivial proofs.