I heard that it is still an open problem whether $Aut(F_n), n\geq 4$ has Kazhdan's property (T), where $F_n$ denotes the non-abelian free group on $n$ generators, my question is:
Does there exist any infinite subgroup $H$ of $Aut(F_n), ~3<n\leq \infty$ such that $H$ has Kazhdan's property (T)?
$\DeclareMathOperator\Aut{Aut}$It's open too. Indeed, the question whether $\Aut(F_n)$ has the Haagerup Property is open as well (actually for all $n\ge 2$). So this one (for $n\ge 4$) is just intermediate between the two.
[For an arbitrary infinite discrete group $G$ we have implications:
$G$ has Kazhdan's T $\Rightarrow$ $G$ has an infinite subgroup with Kazhdan's T $\Rightarrow$ $G$ has an infinite subgroup with relative Kazhdan's T in $G$ $\Rightarrow$ $G$ does not have Haagerup Property.
None of the converses holds. For $n\ge 4$ all are unknown for $\Aut(F_n)$. For $\Aut(F_3)$ the first two fail, and the next two are open, and for $\Aut(F_2)$ (as well as $\operatorname{Out}(F_3)$ and the braid groups $B_{n\ge 4}$) the first three fail and the last is open.]
Update: it has been proved since then that $\Aut(F_n)$ has Property T for $n=5$ (Kaluba, Nowak, Ozawa: arXiv link, arXiv 2017, Math. Ann. 2019) and $n\ge 6$ (Kaluba, Kielak, Nowak: arXiv link, arXiv 2018, Annals 2021). Unless I have missed something, the case $n=4$ remains open.