Does anybody know the answer to this or a good way to go about working this out? I have a list for $GL_2(Z/pZ)$ and I am trying to lift it to this; I have mostly been using fairly elementary algebraic methods. Thank you.

$\begingroup$ Steven Sam has a blog post about this at: concretenonsense.wordpress.com/2009/09/14/… $\endgroup$ – Gwyn Whieldon Mar 11 '14 at 13:44

1$\begingroup$ Note that for this it is really important that it is not just the cyclic group of order $p^2$. It that group with the usual structure of a ring. $\endgroup$ – Tobias Kildetoft Mar 11 '14 at 13:58

$\begingroup$ @Tobias, I edited the title to reflect your comment. $\endgroup$ – Nick Gill Mar 11 '14 at 15:17

$\begingroup$ @Gwyn, I think Steven Sam's blogpost only covers conjugacy classes for $GL_2(q)$, i.e. over a finite field. It seems like the MO knows this theory, but is unclear how to lift to the ring $Z/p^2Z$.... I might be wrong! $\endgroup$ – Nick Gill Mar 11 '14 at 15:27

$\begingroup$ Ah, good point. I misread what he was asking for. $\endgroup$ – Gwyn Whieldon Mar 11 '14 at 15:33
I don't have time to give a full answer now, but here's how I would try and answer this question:
You first need to establish what the normal subgroups of $GL_2(\mathbb{Z}/p^2\mathbb{Z})$ are. This is done in a fair amount of detail at this MO question (at least for $SL_2$, and the same method will generalize to $GL_2$).
Now, as explained in the linked answer above, $G=GL_2(\mathbb{Z}/p^2\mathbb{Z})$ has a normal subgroup $N$ of order $p^3$ which can be thought of as a 3dimensional module for the group $G/N\cong GL_2(\mathbb{Z}/p\mathbb{Z})$. To get the fulllist of conjugacy classes, one needs to examine each coset $gN$ in turn, considering it as a module for the group $C_{G/N}(gN)$.