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

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.

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).