Strongly intersection-closed not implies invariance

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup metaproperties. That is, it states that every subgroup satisfying the first subgroup metaproperty (i.e., strongly intersection-closed subgroup property) need not satisfy the second subgroup metaproperty (i.e., invariance property)
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Statement

General statement

It is possible to have a strongly intersection-closed subgroup property $p$ (a subgroup property closed under arbitrary intersections, including the empty intersection, and therefore true for every group as a subgroup of itself) that is not an invariance property.

Statement with symbols using a specific group

It is possible to have a strongly intersection-closed subgroup property $p$ and a group $G$ such that it is not possible to find a collection $F$ of functions from $G$ to itself for which the set of subgroups invariant under $F$ is precisely the set of subgroups satisfying $p$.

Facts used

1. Invariance implies union-closed
2. Strongly intersection-closed not implies union-closed

Proof

The proof follows directly by combining Facts (1) and (2). Specifically, the counterexample for (2) can be used as a counterexample.