If $H$ is a subgroup of a finite group $G$, let's say that $H$ has the uniform intersection property in $G$ if for any collection $H_1, H_2, \ldots, H_k$ of $k$ distinct conjugates of $H$ in $G$, the cardinality of the intersection $H_1\cap H_2\cap\cdots\cap H_k$ depends only on $k$. (In other words, the size of the intersection is independent of the particular choice of $k$ conjugates of $H$.)
For example, any normal subgroup has this property (vacuously), as does any subgroup of prime order.
Now define a finite group $G$ to have the uniform maximal intersection property (for short, $G$ is/has UMIP) if each of its maximal subgroups has the uniform intersection property in $G$.
Nilpotent finite groups are UMIP as they lack non-normal maximal subgroups. The smallest group that is not UMIP is the alternating group $A_5$ of degree $5$. The smallest soluble group that is not UMIP is the group $(S_3\times S_3)\rtimes C_2$ of order $72$, where the cyclic group $C_2$ of order $2$ acts by interchanging the two copies of the symmetric group $S_3$. I think it might be the case that supersoluble groups are UMIP, but I'm not quite there yet. (However, the group $A_4$ shows that UMIP groups need not be supersoluble.)
Question: Do we know which groups are UMIP? Has this property been studied before? (Almost certainly under a different name than the one I made up.)
Motivation. It might be worth mentioning where this strange notion came from. I recently had a need to try to show that a maximal subgroup $M$ of a group $G$ was normal (under a cocktail of assumptions) and hoped to try to show that if it was not, then the union $\bigcup_{g\in G}M^g$ of its conjugates would be "too big" in the sense that $\left|\bigcup_{g\in G}M^g\right|\geq \left|G\right|$. So, I wanted to try to get a handle on the number of elements in that union by using the set-theoretic version of inclusion-exclusion: $$\left|M_1\cup M_2\cup\cdots\cup M_m\right| = \sum_{k=1}^m (-1)^{k+1}\left(\sum_{1\leq j_1<j_2<\cdots < j_k\leq m}\left|M_{j_1}\cap\cdots\cap M_{j_k}\right|\right).$$ (Here, $M_1,\ldots,M_m$, with $m = [G:M],$ are the conjugates of $M$ in $G$.) This idea failed because I wasn't able to get a handle on the sizes of those intersections appearing in the sums, and I got distracted from my original problem thinking about for which groups the problem would simplify by having the orders of those intersections all be the same for a given number of conjugates.