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.