Is a finite group given by its character table if its Sylow subgroups are so?

Question

As pointed out by Mikko Korhonen in this answer, Özdem Çelik proved (in 1976 here) that a finite group whose Sylow subgroups are cyclic (called a Z-group) is determined by its character table.

Now there are many results and conjectures relating character tables and Sylow subgroups (see this paper of Gabriel Navarro), the most famous being perhaps the McKay conjecture.

This leads to wonder whether Çelik's theorem can be extended*.

Question 1: Is a finite group determined by its character table only if its Sylow subgroups are so?
Answer (Alex B.): No.

Question 2: Is a finite group not in a Brauer pair only if its Sylow subgroups are so?
(negative answer suspected by Alex B.)

*Question 3: Is a finite group determined by its character table if its Sylow subgroups are so?
(it is this question which wonders whether Çelik's theorem can be extended)

Question 4: Is a finite group not in a Brauer pair if its Sylow subgroups are so?

Answer 1

The answer to the first question is negative. The group ${\rm SL}_2(\mathbb{F}_3)$ has a $2$-Sylow subgroup isomorphic to $Q_8$, which is not determined by its character table, but ${\rm SL}_2(\mathbb{F}_3)$ is the only group with its character table.

I strongly suspect that the answer to the second question is also negative, and it should be not too hard to check this computationally, using the following strategy, but I have not attempted this.

One way of seeing how the counterexample above worked is this: the group $Q_8$ has an outer automoprhism of order $3$ (cyclicly permuting $i$, $j$, and $k$), and ${\rm SL}_2(\mathbb{F}_3)$ is a semidirect product of $Q_8$ and $C_3$, with the action given by this automorphism. On the other hand, $D_8$ has no automorphisms of order $3$, so one cannot produce an analogous construction with this cousin of $Q_8$. That makes it "unlikely" (in no precise sense) that there is another group with the same character table as ${\rm SL}_2(\mathbb{F}_3)$.

I have checked that the group that Lux and Pahlings call $G_{3378}$ in Theorem 2.6.2 has an automorphism $\alpha$ of order $3$, while the other group in the Brauer pair, $G_{3380}$, does not. So I would conjecture that the group $G_{3378}\rtimes\langle \alpha\rangle$ is determined by its character table. This is a group of order $768$, so magma has the complete list of all groups of this order, and the conjecture should not be too hard to check, but I have not done that.