Let $G=(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$, where $p,q>3$ are distinct primes.
$\phi: \mathbb{Z}_q \rightarrow {\rm Aut}(\mathbb{Z}_p \times \mathbb{Z}_p)$.
Suppose $G$ is generated by $s$ and $g$, $G=\langle s,t\rangle$, where $|s|=q, |t|=p$. Let $\phi_s(t)= s^{-1}ts$ and suppose $\{t, u\}$ is a generating set of $\mathbb{Z}_p \times \mathbb{Z}_p$ ($|u|=p$).
In a computation, a person, say $A$ obtain the values $\phi_{s^m}(t)=A_1$ and $\phi_{s^m}(u)=A_2$ for some $s^m \in \mathbb{Z}_q$ and tell $B$ only the values $\{A_1, A_2, p, q\}$.
Then only by using the values $\{A_1, A_2, p, q\}$ it is difficult for $B$ to obtain an idea about the original $\phi$ nor about $s,t,u$ elements even if he knows that $A$ did the computations in some semidirect product of the form $(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_q$, right?
If $B$ is going to check by computing values for all possibilities present,
When $q\mid p+1$ there is a unique automorphism, and if $q\mid p-1$ there are $\frac{q+3}{2}$ automorphisms. Then there are $\frac{q+3}{2}+1$ possibilities to check.
And there are $q-1$ possibilities for $s^m$ (If we don't consider the element $0 \in \mathbb{Z}_q$ which is the additive identity).
For $t$ and $u$ how many possibilities (say $X$ number of pairs)?
So if $B$ is going to check all possibilities he can write an algorithm which will run in polynomial time I think? (($\frac{q+3}{2}+1)(q-1)X$). Am I right?
Argument 2:
But however, if $A$ computes $\phi_{s^m}$ of several more group elements, say $g_1, g_2, \cdots , g_i$, in addition to $\phi_{s^m}(t), \phi_{s^m}(u)$ and gives $B$ the set of values $\{ \phi_{s^m}(t), \phi_{s^m}(u), \phi_{s^m}( g_1), \phi_{s^m}( g_2), \cdots, \phi_{s^m}( g_i)\} $, then $B$ doesn’t know to what values the generating elements $t$ and $u$ are mapped. Then will it be impossible for $B$ to find $\phi, s,t,u $ used by $A$, even if $B$ checks all the possible possibilities ($B$ can check for the possible $s,t,u$ and possible $g_1, g_2, \cdots, g_i$?)?
I thought the following in support of the above argument 2:
Because now other group elements $g_1, g_2, \cdots , g_i$ are also considered, even if $\phi_{s^m}(t) = A_1$, there will be a $g_k$ such that $\phi_{s^w}(g_k)=A_1$, in the above list of values $B$ has received for some $s^w \in \mathbb{Z}_q$. So it becomes impossible to check by going through all the possibilities, since we don’t know which value in the list of values received by $B$ will correspond to the generating elements $t,u$.
Please kindly mention what you think about these arguments.
Thanks a lot in advance.