No, there are no such $\mathcal U$ and $\varphi$.
Let me start the proof with two simplifying observations. First, $\varphi$ is irrelevant, because bijections of $\omega$ to itself preserve selectivity of ultrafilters.
Second, every selective ultrafilter $\mathcal U$ is a P-point, which means that, given any countably many sets $A_n\in\mathcal U$, there is a set $B\in\mathcal U$ almost included in all of the $A_n$"s, i.e., $B-A_n$ is finite for all $n$. To prove that selective ultrafilters $\mathcal U$ have this property, let sets $A_n\in\mathcal U$ be given. If $\bigcap_nA_n\in\mathcal U$, then this intersection serves as the required $B$, so assume $\bigcap_nA_n\notin\mathcal U$. Then we can partition $\omega$ into the pieces $\omega-A_0, A_0-A_1, A_1-A_2, A_2-A_3, \dots$ and $\bigcap_nA_n$, none of which are in $\mathcal U$. Selectivity provides a set $B\in\mathcal U$ that intersects each of these pieces in at most one point. Then $B-A_n$ is finite (in fact it has at most $n+1$ elements) for each $n$, as required.
So now it suffices to show that no P-point ultrafilter $\mathcal U$ can include the density-$1$ filter $\mathcal F_I$. For each natural number $n\geq 2$, partition $\omega$ into $n$ sets each of which has density $\frac1n$; for example, take the congruence classes modulo $n$. Being an ultrafilter, $\mathcal U$ must contain one of these $n$ sets, say $A_n$. If $\mathcal U$ is a P-point, it contains a set $B$ almost included in all of these $A_n$'s. But from almost inclusion it easily follows that $B$ has upper density at most the density $\frac1n$ of $A_n$. (Upper density is defined like density but with $\limsup$ instead of $\lim$.) So $B$ has upper density $0$, which means it has density $0$. Thus, $\omega-B$ has density $1$ and is therefore in $\mathcal F_I$, but it is not in $\mathcal U$ because $B\in\mathcal U$.