While practising on paper I've realized of a property of multiplicative group of integers mod $n$.
First, let's define $G$ being $p$ a prime and $g$ a primitive root mod n or a generator of a subgroup of $p$ whose order is a factor of $|G|$.
Example:
$p=23$
$|G|=p-1=22$
Then find a valid $g$ whose $Ord_{p}(g)=p-1$ or $Ord_{p}(g)=2$ or $Ord_{p}(g)=11$ being $11$ and $2$ the factors of $22$. We'll define $q$ as a factor of $p$ so $q=11$ in this case.
Using the mechanism of testing a primitive root we find that:
$2^{11}$ $mod$ $23 = 1$ So $Ord_{p}(2)=11$
We found that $2$ isn't a primitive root mod n so it generates a subgroup of order $11$
Now we compute $g^{q-1}$ $mod$ $p$:
$2^{10}$ $mod$ $23 = 12$
Now if $Ord_{p}(2)=Ord_{p}(12)$ then:
$12^{10}$ $mod$ $23 = 2$
Summarizing, this property satisfies that if two generators have the same multiplicative order then if you raise the first generator to $p-2$ or $q-1$ you will obtain the second generator as a congruence.
I want to know if this make sense. Would this method reveal any info of the exponent used by the involved parties?
Thanks for your effort and time!