Does $\mathrm{SL}_{n}(\mathbb{Z}/p^{2})$ have the same number of conjugacy classes as $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$? 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 $\mathrm{SL}_{n}(\mathbb{F}_{p}[t]/t^{2})$?

When $p$ does not divide $n$ this follows from a theorem of P. Singla (see this paper). Note that the case when $p$ divides $n$ in this paper has a gap (see Section 5 here). 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.
 A: Oh, I did not know about this ongoing discussion on math overflow. Amri pointed out to me about this discussion today morning only. As I discussed with you in a private communication, I don't know how to fix this at the moment. Moreover in this recent article of mine with M Hassain, we show that  $SL_n$ for $p \mid n$ even for $n=2$ behaves pretty differently as compared to $GL_n$ (see Theorem 1.2). For example Corollary 1.3 of this article shows that the complex group algebras of $SL_2(Z/2^{2r} Z)$ are not isomorphic to $SL_2(F_2[t]/(t^{2r}) )$ for any $r > 1$. This is weaker than conjugacy class question for such groups, if one is interested in that, but still quite interesting given that corresponding group algebra of $GL_2$ are isomorphic. 
A: 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 (arXiv link here).
