In seek of a (counter)example

Question

I am looking for an example of a finite group $G$ and a subgroup $H$ of $G$ with the following property:

The number of conjugacy classes of subgroups of $G$ which are isomorphic to $N_G(H)$ is strictly larger than the number of conjugacy classes of subgroups of $G$ which are isomorphic to $H$.

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I've figured out how to check for this property with GAP, so I'll paste my code below in case anyone finds it useful. (The bit that was giving me trouble was the sixth line of the code.)

# given a finite group G, compute the number of conjugacy classes 
# of subgroups of G which are isomorphic to a given subgroup H.
nccs_isom := function(G,H)
local ccs, K, count;
ccs:=List(ConjugacyClassesSubgroups(G),Representative);
count := 0;
for K in ccs do
  if IdSmallGroup(K) = IdSmallGroup(H) then
    count := count+1;
  fi;
od;
return count;
end;

grps:=AllSmallGroups([1..100],IsAbelian,false);;

# the actual search
for g in grps do
  ccs:=List(ConjugacyClassesSubgroups(g),Representative);
  for h in ccs do
    norm:=Normalizer(g,h);
    if nccs_isom(g,norm) > nccs_isom(g,h) then  
      Print(IdSmallGroup(g),"  ",IdSmallGroup(h),"\n");                              
    fi;
  od;
od;

Answer 1

By brute force computer search, I found a counterexample $G=\mathtt{SmallGroup}(64,33)$.

For $H$ take a cyclic subgroup of order $4$ with normalizer of order $8$. There is only one class of such subgroups, and $N_G(H) \cong C_4 \times C_2$.

There are four conjugacy classes of $G$ isomorphic to $H$ and five classes isomorphic to $N_G(H)$.