Diffeomorphism of Lens Spaces Let $p,q \in \mathbb{Z}$, with $p > 0$ and $gcd(p,q) = 1$. The Lens space $L(p,q)$ is defined by $\mathbb{S}^3/\mathbb{Z}_p$ where the action is given by $n \cdot (z_1,z_2) = (e^{2\pi i n /p}z_1,e^{2\pi i n q/p}z_2)$. Here we are identifying $\mathbb{S}^3$ with the unit sphere of $\mathbb{C} \times \mathbb{C}$. I wish to show that if $q + q^\prime\equiv 0 \;(mod\; p)$ or $qq^\prime \equiv 1 \;(mod \; p)$ then $L(p,q)$ is diffeomorphic to $L(p,q^\prime)$. 
I tried defining a map from $\mathbb{S}^3$ to $\mathbb{S}^3$ that could induce such diffeomorphisms but I wasn't able to find it. Any sugestion? 
 A: You can do this by thinking carefully about the orbits of points under the various group actions. I'll look at the case where $q'q\equiv 1 \bmod p$ first.
Under the action of $\mathbb{Z}/p\mathbb{Z}$, given by $(z_1,z_2)\mapsto (z_1e^{2\pi i / p},z_2 e^{2\pi i q / p})$ the orbit of the point $(w_1,w_2)$ is the set
$$O_1=\{(w_1e^{2\pi i k / p},w_2 e^{2\pi i kq / p})\,|\, k=0, \dots, p-1\}.$$
Now we study orbit of $(w_2,w_1)$, the under the action given by 
$(z_1,z_2)\mapsto (z_1e^{2\pi i / p},z_2 e^{2\pi i q' / p})$. This orbit is 
$$O_2=\{(w_2e^{2\pi i k' / p},w_1 e^{2\pi i k'q' / p})\,|\, k'=0, \dots, p-1\},$$
which, by substituting $k'=kq \bmod p$, can be rewritten as 
$$O_2=\{(w_2e^{2\pi i kq / p},w_1 e^{2\pi i k / p})\,|\, k=0, \dots, p-1\}.$$
Thus the map $\phi \colon S^3 \rightarrow S^3$ given by
$\phi(z_1,z_2)=(z_2,z_1)$ carries the orbit $O_1$ of $(w_1,w_2)$ under the first action onto the orbit $O_2$ of $(w_2,w_1)$ under the second action, thus inducing a diffeomorphism between $L(p,q)$ and $L(p,q')$ where $qq'\equiv 1 \bmod p$.
The case of $-q$ is similar. Consider the orbit of $(w_1,\overline{w_2})$, the under the action given by 
$(z_1,z_2)\mapsto (z_1e^{2\pi i / p},z_2 e^{-2\pi i q / p})$. This orbit is 
$$O_3=\{(w_1 e^{2\pi i k' / p}, \overline{w_2} e^{-2\pi i k'q / p})\,|\, k'=0, \dots, p-1\}.$$
Thus $O_3$ is the image of $O_1$ under the map $\psi\colon S^3\rightarrow S^3$ given by $\psi(z_1,z_2)=(z_1,\overline{z_2})$. Thus $\psi$ induces a diffeomorphism between $L(p,q)$ and $L(p,-q)$. Here I'm using $\overline{z}$ to denote complex conjugation.
