TITLE: Bijection between derangements and good permutations
QUESTION [5 upvotes]: A good permutation is a permutation of the numbers $1$ to $n$, such that $i$ is not followed by $i+1$ at any position in the permutation, for any $i \in \left\{1,2,\ldots,n-1\right\}$. Call the number of such permutations $S(n)$.
Also, $D(n)$ is the number of derangements of $\left\{1,2,\ldots,n\right\}$.
I am required to show that $S(n) = D(n) + D(n-1)$.

Note that an algebraic solution is possible, but I need a combinatorial argument i.e. by showing bijections between the two sets.

REPLY [3 votes]: This does not in fact work; see the comments. Then see the answers to this question.
HINT: Let $\pi=\pi_1\pi_2\dots\pi_n$ be a permutation of $[n]=\{1,\dots,n\}$. Define a map 
$$\hat\pi:[n]\to[n]:\pi_k\mapsto\begin{cases}
\pi_{k+1}-1,&\text{if }k<n\text{ and }\pi_{k+1}\ne 1\\
\pi_1-1,&\text{if }k=n\text{ and }\pi_1\ne 1\\
n,&\text{otherwise}\;;
\end{cases}$$
$\hat\pi$ is a permutation of $[n]$, and in fact the map $\pi\mapsto\hat\pi$ is a bijection on $S_n$, the set of permutations of $[n]$.
Suppose that $\pi$ is good. Then $\hat\pi$ has no fixed point in $[n-1]=\{1,\dots,n-1\}$. Can you fill in the details and finish it from there?