Showing that $Q_n=D_n+D_{n-1}$

Let $[a,b]=\{i\in\mathbf{Z}\mid a\le i\le b\}$ and let $[b]=[1,b].$ Let $S_{a,b}$ be the set of permutations of $[a,b]$ and let $S_n$ be the set of permutations of $[n].$ Let $\overline{S}_{a,b}=\{\sigma\in S_{a,b}\mid \sigma(b)=b\}$ be the set of permutations of $[a,b]$ that fix $b.$ Similarly, Let $\overline{S}_n=\{\sigma\in S_n\mid \sigma(n)=n\}$ be the set of permutations of $[n]$ that fix $n.$

The result will follow by exhibiting a bijective mapping $\rho_n:S_n\to\overline{S}_{n+1}$ that has the property that if $\sigma$ is an element of $S_n,$ $\tau$ is the element of $\overline{S}_{n+1}$ given by $\tau=\rho_n(\sigma),$ $f$ is an element of $[n],$ and $e$ is given by $\tau(e)=f,$ then the property $\tau(e+1)=\tau(e)+1=f+1$ holds if and only if $f$ is a fixed point of $\sigma.$

To define $\rho_n$ we introduce some notation. For distinct elements $i_1,\ldots,i_n\in[n]$ let $\langle i_1i_2\ldots i_n\rangle$ denote the element $\sigma\in S_n$ that satisfies $\sigma(j)=i_j$ for $1\le j\le n.$

We use usual cycle notation to represent swaps, which act from the left. So $(35)\langle123456\rangle=\langle125436\rangle,$ and $(23)(35)\langle123456\rangle=\langle135426\rangle.$ Every element of $S_n$ has a unique swap representation $(a_1b_1)(a_2b_2)\ldots(a_kb_k)\langle12\ldots n\rangle$ where $0\le k\le n-1,$ $a_1<a_2<\ldots<a_k,$ and $a_i<b_i$ for all $1\le i\le k.$ For example, $$ \langle 326451\rangle=(13)\langle 126453\rangle=(13)(36)\langle 123456\rangle. $$ Observe that $f$ is a fixed point of $\sigma\in S_n$ if and only if $f$ does not appear among the $i_1,\ldots,i_k,j_1,\ldots,j_k$ in the swap representation of $\sigma.$

We define the insertion operator $[ai]:\overline{S}_{a+1,n+1}\to\overline{S}_{a,n+1},$ where $a\in[1,n]$ and $i\in[a,n].$ Let $\tau\in\overline{S}_{a+1,n+1}.$ Then $[ai]\tau$ is defined to be the element of $\overline{S}_{a,n+1}$ obtained by inserting $a$ immediately in front of $i+1.$ So, for example, $[24]\langle3546\rangle=\langle32546\rangle.$ Every element of $\overline{S}_{n+1}$ has a unique insertion representation $[1b_1][2b_2]\ldots[nb_n]\langle (n+1)\rangle,$ where $b_j\ge j$ for all $1\le j\le n.$ For example, $$ \begin{aligned} &\langle 1456237\rangle=[13]\langle 456237\rangle=[13][22]\langle 45637\rangle =[13][22][36]\langle 4567\rangle\\ &=[13][22][36][44]\langle 567\rangle=[13][22][36][44][55]\langle 67\rangle=[13][22][36][44][55][66]\langle 7\rangle. \end{aligned} $$ Observe that $a$ immediately precedes $a+1$ in $\tau\in\overline{S}_{n+1}$ if and only if the only occurrence of $a$ in the insertion representation of $\tau$ is of the form $[aa].$

The bijective mapping $\rho_n:S_n\to\overline{S}_{n+1}$ is now defined as follows. Let $\sigma\in S_n$ and compute the swap representation of $\sigma.$ We may add null swaps of the form $(aa)$ so that the swap representation of $\sigma$ takes the form $(1b_1)(2b_2)\ldots(nb_n)\langle 123\ldots n\rangle,$ where $b_j\ge j$ for all $1\le j\le n.$ Then define $\rho_n(\sigma)$ by replacing swaps with insertions and replacing $\langle 123\ldots n\rangle$ with $\langle (n+1)\rangle:$ $$ \rho_n(\sigma)=[1b_1][2b_2]\ldots[nb_n]\langle (n+1)\rangle. $$

Examples: We show $\rho_3:S_3\to\overline{S}_4$ and $\rho_4:S_4\to\overline{S}_5$ explicitly. On the left, dots mark fixed points; on the right, overscores mark sequences of consecutive elements. $$ \begin{aligned} \langle\dot{1}\dot{2}\dot{3}\rangle=(11)(22)(33)\langle123\rangle&\mapsto[11][22][33]\langle4\rangle=\langle\overline{1234}\rangle\\ \langle\dot{1}32\rangle=(11)(23)(33)\langle123\rangle&\mapsto[11][23][33]\langle4\rangle=\langle3\overline{12}4\rangle\\ \langle21\dot{3}\rangle=(12)(22)(33)\langle123\rangle&\mapsto[12][22][33]\langle4\rangle=\langle21\overline{34}\rangle\\ \langle231\rangle=(12)(23)(33)\langle123\rangle&\mapsto[12][23][33]\langle4\rangle=\langle1324\rangle\\ \langle312\rangle=(13)(23)(33)\langle123\rangle&\mapsto[13][23][33]\langle4\rangle=\langle3214\rangle\\ \langle3\dot{2}1\rangle=(13)(22)(33)\langle123\rangle&\mapsto[13][22][33]\langle4\rangle=\langle\overline{23}14\rangle \end{aligned} $$ Observe that the two derangements, $\langle231\rangle$ and $\langle312\rangle,$ have images that contain no sequences of consecutive elements.

$$ \begin{aligned} \langle\dot{1}\dot{2}\dot{3}\dot{4}\rangle=(11)(22)(33)(44)\langle1234\rangle&\mapsto[11][22][33][44]\langle5\rangle=\langle\overline{12345}\rangle\\ \langle\dot{1}\dot{2}43\rangle=(11)(22)(34)(44)\langle1234\rangle&\mapsto[11][22][34][44]\langle5\rangle=\langle4\overline{123}5\rangle\\ \langle\dot{1}32\dot{4}\rangle=(11)(23)(33)(44)\langle1234\rangle&\mapsto[11][23][33][44]\langle5\rangle=\langle3\overline{12}\,\overline{45}\rangle\\ \langle\dot{1}342\rangle=(11)(23)(34)(44)\langle1234\rangle&\mapsto[11][23][34][44]\langle5\rangle=\langle\overline{12}435\rangle\\ \langle\dot{1}423\rangle=(11)(24)(34)(44)\langle1234\rangle&\mapsto[11][24][34][44]\langle5\rangle=\langle43\overline{12}5\rangle\\ \langle\dot{1}4\dot{3}2\rangle=(11)(24)(33)(44)\langle1234\rangle&\mapsto[11][24][33][44]\langle5\rangle=\langle\overline{34}\,\overline{12}5\rangle\\ \langle21\dot{3}\dot{4}\rangle=(12)(22)(33)(44)\langle1234\rangle&\mapsto[12][22][33][44]\langle5\rangle=\langle21\overline{345}\rangle\\ \langle2143\rangle=(12)(22)(34)(44)\langle1234\rangle&\mapsto[12][22][34][44]\langle5\rangle=\langle42135\rangle\\ \langle231\dot{4}\rangle=(12)(23)(33)(44)\langle1234\rangle&\mapsto[12][23][33][44]\langle5\rangle=\langle132\overline{45}\rangle\\ \langle2341\rangle=(12)(23)(34)(44)\langle1234\rangle&\mapsto[12][23][34][44]\langle5\rangle=\langle24135\rangle\\ \langle2413\rangle=(12)(24)(34)(44)\langle1234\rangle&\mapsto[12][24][34][44]\langle5\rangle=\langle41325\rangle\\ \langle24\dot{3}1\rangle=(12)(24)(33)(44)\langle1234\rangle&\mapsto[12][24][33][44]\langle5\rangle=\langle1\overline{34}24\rangle\\ \langle312\dot{4}\rangle=(13)(23)(33)(44)\langle1234\rangle&\mapsto[13][23][33][44]\langle5\rangle=\langle321\overline{45}\rangle\\ \langle3142\rangle=(13)(23)(34)(44)\langle1234\rangle&\mapsto[13][23][34][44]\langle5\rangle=\langle21435\rangle\\ \langle3\dot{2}1\dot{4}\rangle=(13)(22)(33)(44)\langle1234\rangle&\mapsto[13][22][33][44]\langle5\rangle=\langle\overline{23}1\overline{45}\rangle\\ \langle3\dot{2}41\rangle=(13)(22)(34)(44)\langle1234\rangle&\mapsto[13][22][34][44]\langle5\rangle=\langle14\overline{23}5\rangle\\ \langle3412\rangle=(13)(24)(33)(44)\langle1234\rangle&\mapsto[13][24][33][44]\langle5\rangle=\langle31425\rangle\\ \langle3421\rangle=(13)(24)(34)(44)\langle1234\rangle&\mapsto[13][24][34][44]\langle5\rangle=\langle14325\rangle\\ \langle4123\rangle=(14)(24)(34)(44)\langle1234\rangle&\mapsto[14][24][34][44]\langle5\rangle=\langle43215\rangle\\ \langle41\dot{3}2\rangle=(14)(24)(33)(44)\langle1234\rangle&\mapsto[14][24][33][44]\langle5\rangle=\langle\overline{34}215\rangle\\ \langle4\dot{2}13\rangle=(14)(22)(34)(44)\langle1234\rangle&\mapsto[14][22][34][44]\langle5\rangle=\langle4\overline{23}15\rangle\\ \langle4\dot{2}\dot{3}1\rangle=(14)(22)(33)(44)\langle1234\rangle&\mapsto[14][22][33][44]\langle5\rangle=\langle\overline{234}15\rangle\\ \langle4312\rangle=(14)(23)(34)(44)\langle1234\rangle&\mapsto[14][23][34][44]\langle5\rangle=\langle24315\rangle\\ \langle4321\rangle=(14)(23)(33)(44)\langle1234\rangle&\mapsto[14][23][33][44]\langle5\rangle=\langle32415\rangle \end{aligned} $$ Observe that the images of the nine derangements contain no sequences of consecutive elements.

Since there is a one-to-one correspondence between the set of derangements of $[n]$ and the set of permutations of $[n+1]$ that fix $n+1$ and contain no sequences of consecutive elements (we let $\overline{T}_{n+1}$ denote this set), we have $$ D_n=\lvert \overline{T}_{n+1}\rvert. $$ But there is an immediate bijection between $\overline{T}_{n+1}$ and $T_n-\overline{T}_n,$ the set of elements of $T_n$ that do not fix $n.$ Therefore $$ D_n=\lvert T_n-\overline{T}_n\rvert $$ and $$ D_n+D_{n-1}=\lvert T_n-\overline{T}_n\rvert+\lvert \overline{T}_n\rvert=\lvert T_n\rvert. $$

(Note: this answer does not contain a bijection between the two sets, which is what the OP was originally hoping for, but it does show that the two counting problems are structurally equivalent and therefore have the same answer.)

The belief stated in the last paragraph of your question is correct: we can show that the set $\mathcal{D}_{n-1}$ of derangements of $\{1,2,3,\ldots,n-1\}$ and the set of permutations $\sigma$ of $\{1,2,3,\ldots,n\}$ with the properties

• $\sigma(n)=n$,
• $\sigma(i+1)\ne\sigma(i)+1,$ for $1\le i\le n-1,$

are equinumerous. That is, $\mathcal{D}_{n-1}$ and the set of elements of $T_n$ that fix $n$ have the same size. Call the latter set $\overline{T}_n.$

This result will prove the equation in your title since there is an obvious bijection between the set of elements of $T_n$ that do not fix $n$ and the set $\overline{T}_{n+1}$ of elements of $T_{n+1}$ that do fix $n+1,$ which will then be equinumerous with $\mathcal{D}_n.$

To enumerate $\mathcal{D}_{n-1}$, let $F=S_{n-1}$ be the set of permutations of $1,2,\ldots,n-1.$ Let $F_j$ be the set of elements of $F$ that fix $j,$ that is, elements $\sigma$ such that $\sigma(j)=j.$ In general, let $F_{ijk\ldots}=F_i\cap F_j\cap F_k\cap\ldots$ be the set of elements that fix $i,$ $j,$ $k,\ldots$, that is, elements $\sigma$ such that $\sigma(i)=i,$ $\sigma(j)=j,$ $\sigma(k)=k,\ldots$ Observe that $\lvert F_{i_1i_2\ldots i_k}\rvert=(n-1-k)!$ since only $n-1-k$ elements are free to move. Since the derangements are those elements that fix no element, the principle of inclusion-exclusion gives $$\begin{aligned}\lvert\mathcal{D}_{n-1}\rvert&=\lvert F\rvert-\sum_{i=1}^{n-1}\lvert F_i\rvert+\sum_{1\le i<j\le n-1}\lvert F_{ij}\rvert-\ldots\\ &=(n-1)!-\binom{n-1}{1}(n-2)!+\binom{n-1}{2}(n-3)!-\ldots\end{aligned}$$

To enumerate $\overline{T}_n$, let $G$ be the set of permutations of $1,2,\ldots,n$ that fix $n.$ Let $G_i,$ $1\le i\le n-1,$ be the subset of $G$ consisting of elements $\sigma$ such that $\sigma(i+1)=\sigma(i)+1.$ In general, let $G_{ijk\ldots}=G_i\cap G_j\cap G_k\cap\ldots$ be the set of elements of $G$ such that $\sigma(i+1)=\sigma(i)+1,$ $\sigma(j+1)=\sigma(j)+1,$ $\sigma(k+1)=\sigma(k)+1,\ldots$ We claim that, once again, $\lvert G_{i_1i_2\ldots i_k}\rvert=(n-1-k)!.$ This follows by noting that every constraint $\sigma(i_j+1)=\sigma(i_j)+1$ reduces the number of elements that can move independently by $1,$ leaving only $n-1-k$ elements that are free to move. One can imagine element $i$ becoming "glued" to element $i+1,$ so that they must move as a block.

For example, let $n=9$ and consider $G_{12478}.$ Since $8$ must immediately precede $9,$ which is fixed, and $7$ must immediately precede $8,$ which is now fixed as well, the elements $789$ are fixed. At the same time, $2$ must immediately precede $3$ and $1$ must immediately precede $2,$ meaning that the string $123$ can only move as a block. Similarly, the string $45$ can only move as a block. As a consequence the number of permutations in $G_{12478}$ is the number of ways of permuting the "objects" $123,$ $45,$ and $6,$ with the object $789$ fixed in place. So $G_{12478}=3!=(9-1-5)!,$ in agreement with the claim.

Since $\lvert G_{ijk\ldots}\rvert=\lvert F_{ijk\ldots}\rvert$ for all choices of subscripts, the principle of inclusion-exclusion implies that the computation of $\lvert\overline{T}_n\rvert$ is identical to that of $\lvert\mathcal{D}_{n-1}\rvert,$ so that they have the same final value: $$\begin{aligned}\lvert\overline{T}_n\rvert&=\lvert G\rvert-\sum_{i=1}^{n-1}\lvert G_i\rvert+\sum_{1\le i<j\le n-1}\lvert G_{ij}\rvert-\ldots\\ &=(n-1)!-\binom{n-1}{1}(n-2)!+\binom{n-1}{2}(n-3)!-\ldots\end{aligned}$$

EDIT: What's below is actually WRONG, due to multiple issues noted in the comments (the equation taken modulo $n$ leads to situations where a consecutive $n1$ causes problems, and it's not necessarily a bijection.

I think a variant on dkuper's argument can be made to work.

If I understand correctly, the permutations in $Q_n$ are those which satisfy $\sigma(i)+1 \neq \sigma(i+1)$ for $1 \leq i \leq n-1$. Conversely, we can think of $D_n+D_{n-1}$ as consisting of those permutations satisfying $\tau(i) \neq i$ for $1 \leq i \leq n-1$ ($D_n$ corresponds to those permutations with $\tau(n) \neq n$ as well, while $D_{n-1}$ corresponds to fixing $\tau(n)=n$).

This suggests that we construct our bijection in such a way that $\sigma(i)+1=\sigma(i+1)$ if and only if $\tau(i)=i$. The former equation can be rewritten as $i=\sigma^{-1} (\sigma(i+1)-1)$. So if we define our bijection by taking $\sigma$ to the permutation satisfying $$\tau(i)=\sigma^{-1}\left(\sigma(i+1)-1\right),$$ where addition and subtraction are taken modulo $n$, things work the way we want them to.