For the concrete case of the cyclic group $C_p$ acting linearly on $S^3$, there's a very explicit construction. Call $\omega = e^{2\pi i/p}$. Fix an integer $q$ coprime with $p$, and let us look at the action $\lambda_q$ on $\mathbb{C}^2$ generated by the diagonal matrix with entries $\omega, \omega^q$. $\lambda_q$ restricts to an action (which I will still call $\lambda_q$) on $S^3$ (as the unit sphere in $\mathbb C^2$ whose quotient is $L(p,q)$.
Take $M = V(x_0^p + \dots + x_3^p)$, the Fermat hypersurface of degree $d$ in $\mathbb{CP}^3$. On $\mathbb{CP}^3$ we define two actions $\psi$ and $\phi$ of $C_p$. Calling $g$ the generator of $C_p$, the two actions are given by:
$\phi_g(x_0:x_1:x_2:x_3) = (\omega x_0 : \omega^{-q}x_1 : x_2 : x_3)$, and $\psi_g(x_0:x_1:x_2:x_3) = (x_0 : x_1 : \omega x_2 : x_3)$.
I claim that:
- Both actions preserve $M$.
- The fixed point set of the action $\phi$ on $M$ is the set of points $(0:0:1:\omega^a)$ as $a$ varies, and the action is semi-free (i.e. there are no new fixed points appearing when you take powers of $g$).
- $\psi_g$ cyclically permutes the fixed points of $\phi_g$ on $M$.
- $\phi_g$ and $\psi_g$ commute.
- The linearised action of $\phi_g$ on $M$ at $(0:0:1:1)$ is $\lambda_{-q}$.
I will not justify these points (the only one that maybe requires some care is the last one), and I will just take them for granted. Now take a small $\phi_g$-invariant ball $B$ in $M$ centred at $(0:0:1:1)$ and remove all its $\psi_g$-orbit in $M$, to get $M_0$. The action of $\phi$ on $M_0$ is free (because $\phi_{g^k}$ has the same fixed points as $\phi_g$ for each $0 < k< p$) and it extends the linear action $\lambda_q$ on $pS^3 = \sqcup_k \psi_{g^k} \partial B$. (Note that there is a $-q$ in an exponent of $\omega$ when defining $\phi$: this is because the boundary of $M_0$ is the boundary of $B$ with its orientation reversed.)
