Homeomorphisms between lens spaces and Poincaré conjecture It is known that $L(p,q)\cong L(p,q')$ if and only if $q\equiv \pm (q')^{\pm 1}\mod p$ where
$$
L(p,q)\cong{\mathbb{S}^3}/\mathord{\sim}
$$
is the quotient space generated by the $\mathbb{Z}_p$-action
$$
\rho(z_1,z_2)=(\zeta z_1,\zeta^q z_2)
$$
where $\zeta$ is a pth root of unity and $p$ and $q$ are coprimes.
Right to left implication can be proven using Reidemeister torsion (https://www3.nd.edu/~lnicolae/Torsion.pdf page 104). However, left to right proof is supposed to be verified easily by the reader. Does someone know how to construct the homeomorphism between $L(p,q)$ and $L(p,q')$ assuming the congruence on $q$ and $q'$?
EXTRA QUESTION:
Does the Poincaré conjecture imply that every closed compact orientable 3-manifold with $\pi_1(X)\cong \mathbb{Z}_p$ is a lens space?
 A: For two roots of unity $\lambda,\mu$ define $\rho_{\lambda,\mu}: (z,w)\mapsto (\lambda z, \mu w)$. Then $\rho_{\lambda,\mu}$ and $\rho_{\lambda,\bar\mu}$ are conjugate via the orthogonal map
$$
(z,w)\mapsto (z, \bar{w}). 
$$
This shows that $L(p,q)$ is homeomorphic to $L(p,-q)$.
Next, if $q$ is coprime to the orders of $\lambda,\mu$ then the subgroup generated by $\rho_{\lambda,\mu}$ is the same as the one generated by $(\rho_{\lambda,\mu})^q$.
Lastly, assuming that $p$ is coprime to $q$, $q'$,
$qq'\equiv 1$  mod $p$, and
$$
\lambda=\exp(i 2\pi/p), \mu= \exp(i 2\pi q/p), 
\mu'= \exp(i 2\pi q'/p), 
$$
then
$$
\lambda^{q'}= \exp(i 2\pi q'/p), 
$$
$$
\mu^{q'}= \exp(i 2\pi /p).  
$$
In other words, the subgroup generated by $\rho_{\lambda,\mu}$ is the same as generated by
$\rho_{\mu',\lambda}$. However, $\rho_{\mu',\lambda}$ is
conjugate in $U(2)$ to $\rho_{\lambda,\mu'}$ via
$$
(z,w)\mapsto (w,z). 
$$
This proves that $L(p,q)$ is homeomorphic to $L(p,q')$.
Lastly, for your extra question, PC is not enough, you need SSFC. The later was proven by Perelman in the same paper(s) where he proves the Poincare Conjecture.
