Based on a some similar argument due to Roland Schmidt, I think you can take $6$ pairwise different primes $p_1$,$p_2$,$p_3$,$p_4$,$p_5$,$p_6$ representing $1$,$2$,$3$,$4$,$5$,$6$, in the diagram, satisfying the conditions $p_4$ divides $p_1-1$ and $p_2-1$, ... such that if $P_1$,$P_2$,$P_3$,$P_4$,$P_5$,$P_6$ are groups of orders $p_1$, $p_2$, $p_3$, $p_4$, $p_5$, $p_6$, respectively, then the semidirect product $G$ of the cyclic group $C=P_1 \times P_2 \times P_3$ by the cyclic group $D=P_4 \times P_5 \times P_6$ exists in which the operations of $P_4$ are nontrivial on $P_1$ and $P_2$ but trivial on $P_3$, $P_5$ nontrivial on $P_1$ and $P_3$ but trivial on $P_2$, and $P_6$ nontrivial on $P_2$ and $P_3$ but trivial on $P_1$. Then it can be proved that one gets the lattice of normal subgroups in the diagram.