Generalization of Frobenius groups Frobenius group is a transitive permutation group on a finite set, such that no non-trivial element fixes more than one point and some non-trivial element fixes a point.
In other words, if in a transitive permutation group, each element that fixes a point, fixes exactly one point, and there is at least one such element we call it Frobenius.
Let us define a $t$-Frobenius group to be a transitive permutation group, such that each element that fixes a point fixes exactly $t$-points, and there is at least one such element.
Has this concept been studied already? Perhaps under different name? 
Does $t$-Frobenius group exist for every $t$?
Does every $t$-Frobenius group have a regular subgroup?
 A: Yes, these groups exist for all $t$. To see that, let $G$ be a Frobenius group with complement $H$ of size $tu$ for some $u>1$, where $H$ has a normal subgroup $K$ of order $u$. We could, for example, choose $H$ to be a cyclic group of order $tu$, and we can do that for any $t$ and $u$. Then the image of the permutation representation of $G$ on the cosets of $K$ is a $t$-Frobenius group.
But note that not all examples of $t$-Frobenius groups arise in this way from Frobenius groups. For example, $D_8$ on $4$ points is $2$-Frobenius, but does not arise in this way.
No, they do not all have regular subgroups. With above construction with $G$ Frobenius of degree $5$ and order $20$, and $u =2$, the resulting $2$-Frobenius group has no regular subgroup of order $10$.
More generally, if the subgroup $H$ of $G$ is a trivial intersection set, which means that it is disjoint from its distinct conjugates, then the permutation representation of $G$ on the cosets of $H$ is a $t$-Frobenius group, where $t = |N_G(H):H|$. This gives rise to lots more examples.
But this still does not include all examples. An example not of this form is the simple group ${\rm PSL}(2,7)$ acting on the cosets of a Klein $4$-group, which is a $6$-Frobenius group of degree $42$. This is because all nontrivial elements of the stabilizer are conjugate in $G$.
