I have come across the following problem, which is considered as a mathematically hard problem to solve (one way trapdoor) used for cryptography.
Conjugacy search problem: Let $G$ be a non-abelian group. Let $g,h \in G$ be known such that $h=g^x$. Find $x$ ($g^x=x^{-1} gx$).
This concept can be used for non-abelian groups where conjugacy search is difficult.
Can someone guide me to discuss the difficulty/easiness of the conjugacy search in semidirect product groups $(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_\phi \mathbb{Z}_q$, where $\phi:\mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$, $p,q>2$, $p$ and $q$ are distinct primes.
For an element $(h,k) \in (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_\phi \mathbb{Z}_q$, $(h,k)^{-1}=(\phi_k^{-1}(h^{-1}),k^{-1})$.
Then for $(h_1,k_1), (h_2,k_2) \in (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_\phi \mathbb{Z}_q$, suppose $(h_2,k_2)$ is known and $(h_1,k_1)$ is the unknown value (like $x$ in the above definition),
$(h_1,k_1)^{-1}(h_2,k_2)(h_1,k_1)=(\phi_{k_1}^{-1}(h_1^{-1}),{k_1}^{-1})(h_2,k_2)(h_1,k_1)=(\phi_{k_1}^{-1}({h_1}^{-1})\phi_{k_1}^{-1}(h_2),{k_1}^{-1}k_2)(h_1,k_1)=(\phi_{k_1}^{-1}({h_1}^{-1})\phi_{k_1}^{-1}(h_2)\phi_{k_1^{-1}k_2}(h_1),k_2 )$.
Am I right? Afterwards how should I proceed to argue about the difficulty/easiness of the problem?
Thanks a lot in advance.