Can a non-cyclic group of order $p^2$ act on an orientable surface with no free elements? Firstly, can a group of order $p^2$ for a prime $p>2$ that is not cyclic act on an orientable surface of genus $g>1$? If so, is it possible that it acts so that no element acts freely?
My hope is that the answer to the second of these is no, but I'm struggling to establish it. 
Any thoughts would be great!
 A: *

*Yes, very finite group acts (freely) on a closed oriented surface of some genus $>1$. This follows from the covering theory and the fact every finite group is isomorphic to the quotient group of some surface group. 

*Constructing an action of $G=Z_p\times Z_p$ ($p$ is prime) where no elements acts freely is more difficult. Here is a construction. Start with the fact that for every nontrivial element $g\in G$ has the property that $\phi_g: G/<g>\cong Z_p$. Now, list all these quotients $X_g$ (there are finitely many). Take the (finite union) of these $G$-sets (i.e. sets equipped with a $G$-action) and, hence, this union is again a $G$-set, I will call it $X$. By the construction, each element $g\in G$ has at least $p$ fixed points in $X$ (namely, the entire $X_g$). One can optimize the construction so that each $g\ne 1$ has exactly $p$ fixed points. I will extend the $G$-set $X$ to a larger $G$-space $X'$ by enlarging each point $x\in X$ to a 2-disk $D_x$ centered at $x$ on which the stabilizer $G_x$ of $x$ in $G$ acts as a finite group of rotations. Next, attach 1-handles (i.e. products $I\times I$ to $X'$ in $G$-equivariant fashion, until the result is a connected surface with boundary on which $G$ is again acting. I hope you are familiar with the "handle" terminology, it is very standard in geometric topology, as is the idea of building manifolds by attaching 1-handles to 0-handles, etc: This is exactly what I am doing, with the disks $D_g$ serving as $0$-handles. You do the handle attachment inductively: Start by picking one disk $D_{g_1}$ stabilized by $g_1$ and connect it by 1-handles  to $p$ other disks which belong to a $g_1$-orbit of some other disk $D_{g_2}$. Then continue inductively. You have to be a bit careful with orientation: Fix the orientation on each disk $D_g$ ahead of time and attach 1-handles respecting this orientation. 
On each step, the number of connected components goes down, hence, the process eventually terminates in a compact oriented surface $Y$ with boundary equuipped with a $G$-action. Lastly, you attach (equivariantly) 2-handles (i.e. 2-disks) to the boundary components of $Y$ to obtain your surface $Z$.   
A: I found the following paper which defines what I am looking for as a purely non-free group action:
https://link.springer.com/article/10.1007/s00013-017-1068-6
It contains a construction and well as a lower bound for the genus.
As any group of order $p^2$ is abelian, Theorem 3.4 of the above paper yields the minimal genus of a such a non-cyclic group to be $p(p^2-2p-1)/2+1$.
