The confusion is in the notation. You may be expecting cycle notation, where $(1234)$ means $1$ permutes to $2$, etc. In fact, the notation wolfram is using means that each number permutes to the number in its place. Thus $(1234)$ is the identity permutation.
We could rewrite in cycle notation the group $G_1=\{e,(12),(34),(12)(34)\}$ where $e$ is the identity. Then hopefully it is more clear that the orbit of $1$ is $\{1,2\}$.
On that note, you should know that an element is always in its own orbit since each group contains an identity. Thus you should be suspicious that, as you mentioned, it looks like the orbit of $1$ doesn't contain $1$.
Consider the following recurrence: the desired count $Q_n$ is $1$ for
$n=1$ and $1$ for $n=2.$ For $n\gt 2$ we obtain an admissible
permutation either by placing the value $n$ anywhere at $n-1$ possible
positions of an admissible permutation from $Q_{n-1}$ (this is not $n$
because we may not place $n$ next and to the right of $n-1$) or we
construct a permutation having exactly one pair of consecutive numbers
and place the value $n$ between these two. This can be done by taking
a permutation from $Q_{n-2}$ and replacing one of the $n-2$ values by
a fused pair containing the value and its successor and incrementing
the values that are larger than the first element of the fused pair.
We get the recurrence
$$Q_n = (n-1) Q_{n-1} + (n-2) Q_{n-2}$$
and $Q_1= Q_2= 1.$ This yields the sequence
$$1, 3, 11, 53, 309, 2119, 16687, 148329, 1468457, 16019531, 190899411,
\\ 2467007773, 34361893981, 513137616783,\ldots$$
which is OEIS A000255, where a detailed
entry may be found.
The Maple code for this was as follows.
with(combinat);
C :=
proc(n)
option remember;
local perm, pos, res;
res := 0;
perm := firstperm(n);
while type(perm, `list`) do
for pos to n-1 do
if perm[pos] + 1 = perm[pos+1] then
break;
fi;
od;
if pos = n then
res := res + 1;
fi;
perm := nextperm(perm);
od;
res;
end;
Q :=
proc(n)
option remember;
if n = 1 or n = 2 then return 1 end if;
(n - 1)*Q(n - 1) + (n - 2)*Q(n - 2)
end;
Addendum. Here is my perspective on the inclusion-exclusion
approach. We take as the underlying partially ordered set the set $P$
of subsets (these are the nodes of the poset) of $\{1,2,\ldots,n-1\}$
where a subset $S\in P$ represents permutations where the
elements of $S$ are next to their successors, plus possibly some other
elements also next to their successors. The partially ordered set is
ordered by set inclusion. To compute the cardinality of the
permutations corresponding to $S$ suppose that the elements of $S$
listed in order form $m$ blocks. We first remove these from $[n].$ We
must also remove the elements that are consecutive with the rightmost
element of each block, so we have now removed $|S|+m$ elements. We
then put the augmented and fused blocks back into the permutation and
permute them. We have added in $m$ blocks, therefore the net change is
$-|S|-m +m = -|S|.$ Hence by inclusion-exclusion we compute the
quantity
$$\sum_{S\in P, S\ne\emptyset} (-1)^{|S|} (n-|S|)!.$$
Now this depends only on the number $q$ of elements in $S$ so we get
$$\sum_{q=1}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$
We must now ask about the weight assigned to a permutation with $p$
elements (call this set $T$) next to their successor. This permutation
is included in or rather represented by all sets $S$ that are subsets
of the set $T$, which is the poset spanned by the singletons and $T$
being the topmost node. We obtain
$$\sum_{q=1}^p (-1)^q {p\choose q}
= -1 + (1-1)^p = -1.$$
The count assigns the weight minus one to the permutation. It follows
that when we add $n!$ exactly those permutations remain that do not
contain consecutive adjacent values, for a result of
$$n! + \sum_{q=1}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$
We may simplify this to
$$\sum_{q=0}^{n-1} {n-1\choose q} (-1)^q (n-q)!.$$
Best Answer
i) You want all transpositions (ab), and all pairs of disjoint transpositions (ab)(cd).
ii) Assuming that [2143] means, in cycle notation, (12)(34), then A is a subgroup.