Every cyclic p-group is a Sylow p-subgroup of a finite group whose distinct Sylow p-subgroups intersect trivially in pairs (and there is at least one pair).
For instance, let q be a prime congruent to 1 mod p and then the group $\operatorname{AGL}\left(1, q\right)$ is such a group. This is the normalizer of a Sylow q-subgroup in the symmetric group on q points and can be described as the group of affine transformations of a line over the field of q elements:
$\operatorname{AGL}\left(1, q\right) = \left\{ x \mapsto ax + b \mid a, b \in \mathbb{Z}/q\mathbb{Z}, ~~a \ne 0 \right\}.$
Its Sylow p-subgroups are the cyclic subgroups Pb generated by x → zx + b where z is a primitive pth root of unity in Z/qZ. They intersect trivially, since Pb leaves b/(1−z) alone, but moves every other point.
I can handle a few other cases (quaternion, elementary abelian), but I haven't found a (correct) general method. I was a little surprised semi-direct products with faithful (even irreducible) modules was insufficient.
Which p-groups are Sylow p-subgroups of finite groups whose distinct Sylow p-subgroups intersect trivially in pairs, and there is at least one pair?
In other words, though every pair of distinct Sylow p-subgroups of P intersects trivially, it does so vacuously with no pairs. For each p-group P, I want a finite group G with more than one Sylow p-subgroup $P$ where $P\cap P^g$ in $\{ 1, P \}$ for all $g\in G$.