Let $p$ be a prime; $\mathbb{F}\_{p}$ is the field with $p$ elements
and $\mathbb{F}\_{p}[t]$ the ring of polynomials in $t$ over $\mathbb{F}\_{p}$.



> 
> Does $\mathrm{SL}\_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}\_{n}(\mathbb{F}\_{p}[t]/t^{2})$?
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When $p$ does not divide $n$ this follows from a theorem of P. Singla (see [this paper](https://arxiv.org/abs/1101.3696)). Note that the case when $p$ divides $n$ in this paper has a gap (see Section 5 [here](https://arxiv.org/abs/1804.05043)). In fact, when $p$ does not divide $n$, we have the stronger statement that the number of irreducible characters of degree $d$ is the same for both groups, for every $d$. However, we do not know the answer to the question in the title in general when $p$ divides $n$.


One can check that the answer is yes when $p=n=2$ (10 conjugacy classes)
and for $p=n=3$ (127 conjugacy classes), using GAP (the $n=2$ case
can also be done by hand), but for $n=4$, $p=2$ I don't know the
answer, mainly because the only way I know to create the group over
$\mathbb{F}\_{p}[t]/t^{2}$ in GAP is via generators, and this seems
to be very computationally inefficient.



Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$?
I would have preferred to not answer my own question, but here it goes. Yes, the two groups have the same number of conjugacy classes and in fact, the groups $\mathrm{SL}\_{n}(W\_{2}(\mathbb{F}\_{q}))$ and $\mathrm{SL}\_{n}(\mathbb{F}\_{q}[t]/t^{2})$, for $q$ a power of a prime $p$ dividing $n$, have the same number of irreducible characters of dimension $d$ for every integer $d>1$.
This is proved [here](https://www.sciencedirect.com/science/article/pii/S0021869320304701?dgcid=author) (arXiv link [here](https://arxiv.org/abs/1907.09878)).


