Inverting a list using specific rules Consider two lists, one with $n$ integers ordered from $1$ to $n$, and another list with placeholder variables ($x,y$).
We can transform the two lists by taking any two consecutive elements from the first list and without changing the order interchanging them with the elements of the second list.
For example, if $n$ is $8$, we can move from (($1,2,3,4,5,6,7,8$), ($x,y$)) to (($1,x,y,4,5,6,7,8$), ($2,3$)).
For which $n$ does a sequence of moves exist which transforms the starting position (($1,2,...,n$), ($x,y$)) to (($n,n-1,...1$), ($x,y$))?
I can brute force solutions up to n=8:
1: (1, xy)
2: does not exist
3: does not exist
4: does not exist
5: (12345, xy), (xy345, 12), (xy125, 34), (34125, xy), (3xy25, 41), (3xy41, 25), (32541, xy), (xy541, 32), (xy321, 54), (54321, xy)
6: does not exist
7: does not exist
8: (12345678, xy), (xy345678, 12), (xy312678, 45), (xy312458, 67), (xy672458, 31), (xy672431, 58), (xy658431, 72), (xy658721, 43), (xy654321, 87), (87654321, xy)

 A: Instead of treating this as two lists, instead treat it as a single list of length $n+2$, and just continue the numbering as $x=n+1$ and $y=n+2$. By reformulating it in this way, we can rephrase the problem as a problem on permutation groups. In cycle notation, the allowed transformations are permutations of the form $\tau_i = (i \quad n+1)(i+1 \quad n+2)$ for $i \in \{1 \ldots n-1\}$. The problem asks if the permutation $\sigma=(1 \quad n)(2 \quad n-1)(3 \quad n-2)\ldots$ is in the group $H_n$ generated by the $\tau_i$.
Note that each of our generating permutations is even, so $H_n$ is definitely a subgroup of $A_{n+2}$. We claim for $n \ge 5$, $H_n=A_{n+2}$, proving by induction.
For the base case, it is not hard to calculate that $H_5=A_7$ (I used Magma). Assuming $H_{n-1} = A_{n+1}$, consider $H_n$. First note that $H_n$ is transitive on $1 \ldots n+2$...we exhibit by mapping $1$ onto any other element:


*

*$\tau_1 \tau_a$ maps $1$ to $a$ for all $a \in \{1 \ldots n-1\}$

*$\tau_1 \tau_{n-2} \tau_{n-3} \tau_{n-1}$ maps $1$ to $n$

*$\tau_1$ maps $1$ to $n+1$

*$\tau_1 \tau_{n-2} \tau_{n-3}$ maps $1$ to $n+2$


The stabilizer $S$ of the element $n$ in $H_n$ certainly contains all of the permutations $\tau_i$ for $i \in \{1 \ldots n-2\}$, thus $S$ contains a subgroup isomorphic to $H_{n-1} = A_{n+1}$ by the induction hypothesis. Since the orbit of $n$ under $H_n$ has size $n+2$ and the stabilizer of $n$ in $H_n$ has order at least $\frac12(n+1)!$, by the orbit-stabilizer theorem we must have the order of $H_n$ be at least $\frac12(n+2)!$. But $H_n$ is a subgroup of $A_{n+2}$, which has order $\frac12(n+2)!$, forcing equality.
With this proven, the original question comes down to determining if the permutation $\sigma$ is even or odd. It is even when $n \equiv 0,1 \pmod{4}$, since there will be an even number of cycles, and 0 or 1 fixed points, and odd when $n \equiv 2,3\pmod{4}$. Note that we could perform this task for all $n$ if we permitted the second list to end up as $(y,x)$ instead of $(x,y)$.
A: The list $[(1,2,3,\ldots,n-1, n),(x,y)]$ can be reversed just when $n>4$ is a multiple of 4, or one more than a multiple of four.
In the rest of this answer, I'll show that this statement is true, and provide an explicit recipe for reversing any such list. The recipe reverses the list in $9 \lfloor n/4\rfloor + (n \text{ mod } 4) - 1 \,$  steps, providing an upper bound.
$\newcommand{\space}[2]{\underbrace{#2}_{#1}}$

A toolkit of transformation sequences
In this section, we'll describe useful sequences of moves we can
apply to any list. Our notation will be sequences of position pairs
such as 
$$[(1,2), (3,4), (1,2), (2,3), (4, 5), (2, 3), (1, 2), (3, 4), (1,2)].$$
A pair like (4,5) means "Swap the values at positions (4,5) with the
values at positions (n+1, n+2)". They refer to positions in the list,
not values contained in the list.
Here are the tricks we will use:


*

*Reverse the middle 5. The palidromic sequence mentioned above is actually a solution for reversing the list [(1,2,3,4,5), (x, y)], yielding [(5,4,3,2,1),(x,y)].  Note that this strategy is portable: since it only affects those 5 elements, you could apply this sequence to five consecutive elements in any list, and it would reverse them without touching any other elements in the list.
In other words, for any sequence of five elements within a list
$$\ldots \space{A}{a}\space{B}{b}\space{C}{c}\space{D}{d}\space{E}{e} \ldots,$$
the procedure $[(A,B),(C,D), (A,B), (B,C), (D,E),(B,C), (A,B),(C,D), (A,B)]$ reverses that sublist, without touching any elements outside that sublist:
$$\ldots \space{A}{e}\space{B}{d}\space{C}{c}\space{D}{b}\space{E}{a} \ldots.$$
 

*Reverse the middle 4. Similarly, for any sequence of five consecutive elements
$$\ldots \space{A}{a}\space{B}{b}\space{C}{c}\space{D}{d}\space{E}{e} \ldots,$$
you can reverse the first four (ABCD), yielding
$$\ldots \space{A}{d}\space{B}{c}\space{C}{b}\space{D}{a}\space{E}{e} \ldots.$$
The procedure, as you can prove by working out an example, is:
$$[(D,E), (B,C), (A,B), (B,C), (D,E), (A,B), (C,D), (A,B)]$$
This procedure uses the fifth position (E) for temporarily storage, but does not touch any elements outside this sublist.

*Mirror two pairs. For any non-overlapping sublists 
$$\ldots \space{A}{a}\space{B}{b} \ldots \space{C}{c}\space{D}{d}\space{E}{e}\space{F}{f}\space{G}{g}\ldots,$$
you can "mirror" the two pairs AB and FG, yielding 
$$\ldots \space{A}{g}\space{B}{f} \ldots \space{C}{c}\space{D}{d}\space{E}{e}\space{F}{b}\space{G}{a}\ldots.$$
The palindromic procedure is
$$[(A,B), (E,F), (C,D), (D,E), (F,G), (D,E), (C,D), (E,F), (A,B)]$$
This procedure uses positions CDE as temporary storage space but otherwise does not touch any other elements outside these ABCDEFG.
How to construct a solution for any solvable problem
Let $n$ be the number of elements in your list. In this section, we'll see how to use the toolkit methods to reverse any list that can be reversed.


*

*Case $n=4k$: When $n=4$, the problem is unsolvable. When $n$ is any larger multiple of four, the list contains at least five elements and we can reverse the list in the following way.
First, reverse the middle four elements. Then, apply two-pair-mirroring to the two pairs on either side. Continue outward, applying two-pair-mirroring to the successive two pairs on either side. Because the list contains a multiple of four elements, this will eventually terminate successfully with the entire list reversed.
As an example, for $n=12$, the sequence of toolkit applications looks like:

123456789ABC
123487659ABC
12A9876543BC 
CBA987654321 


*Case $n=4k+1$. If $n$ is one more than a multiple of 4, apply the same procedure, but start by reversing the middle five elements.

*Case $n=4k+2$, Case $n=4k+3$. These are the two remaining possibilities, where $n$ is two or three more than a multiple of 4. We'll show in the next section that these cases are provably unsolvable.
Hence we've provded an exhaustive answer: we have a concrete algorithm for reversing the list when $n$ is a multiple of 4 or one more than a multiple of 4, and all other cases are unsolvable.
Unsolvable cases
In this section, we'll show that the remaining cases $n=4k+2$ and $n=4k+3$ cannot be reversed using the allowed transformations.
The proof actually depends on a very general argument about transformations known as the parity theorem. It goes like this:


*

*Any rearrangement of a list can be written as a sequence of swaps (transpositions) $a\leftrightarrow b$. Of course, there will be many different sequences of swaps that result in the same rearrangement.

*Performing the sequence of swaps in reverse undoes the rearrangement.

*One special rearrangement is the identity rearrangement, which does not change the order of elements at all. You can still write it as a various sequences of swaps (which might swap some elements, then swap them back.)

*The identity theorem says that every possible way of writing the identity rearrangment as a sequence of swaps necessarily uses an even number of swaps.

*Using the identity theorem, we can prove a corollary: the parity theorem: suppose you can represent a rearrangement in two ways. The first way uses $k$ swaps, and the second way uses $\ell$ swaps. The parity theorem says that $k$ and $\ell$ are both even or both odd.
The proof of this theorem is that if you do the $k$ swaps then you do the $\ell$ swaps in reverse, the net result is an identity transform. By the identity theorem, this identity transform uses $k+\ell$ swaps so $k+\ell$ must be even, which means that $k$ and $\ell$ are both even or both odd. 
We can use the parity theorem as follows: note that our allowed transformations always perform two swaps at a time. This means that our allowed transformations can only achieve rearrangements with even parity. By the parity theorem, if we show that reversing the list takes an odd number of swaps, it will follow that reversing the list is impossible using our allowed transformations.
Now suppose we have a list like $((1,2,3\,\ldots,n-1, n),(x,y))$ which we want to reverse. We can write this reversal as a sequence of swaps and check whether the number of swaps is odd or even.
Here's a sequence of swaps for reversing the list:
$$\begin{align*}
1 & \leftrightarrow n\\
2 & \leftrightarrow n-1\\
3 & \leftrightarrow n-2\\
& \vdots
 \end{align*}$$
and (x,y) are unchanged. 
Note that there are around $n/2$ swaps in this list. In particular, if $n$ is a multiple of 4, we can group these swaps into pairs with none left over, showing that the number of swaps is even. The same is true if $n$ is one more than a multiple of 4, since the middle element $\lceil n/2\rceil$ stays in place and isn't swapped.
But if $n$ is two more than a multiple of 4, then there's one swap left over when we pair them up — the swap of elements in the middle of this list $a\leftrightarrow a+1$. And if $n$ is three more than a multiple of 4, the swap left over is $a-1 \leftrightarrow a+1$, with the middle element $a$ of the list staying in place.
This shows that it takes an odd number of swaps to reverse the list when $n=4k+2$ or $n=4k+3$. By the parity theorem, every way of reversing the list takes an odd number of swaps. Because our allowed transformations only perform even numbers of swaps, it follows that such reversals are impossible.
