The answer is Yes.

Let me first reformulate the problem. For a given permutation $(a_1,a_2,\dots,a_{2n+1})$, we construct a 2-colored graph $G$ on vertices $[2n+1]:=\{1,2,\dots,2n+1\}$ formed by black edges $\{i,i+1\}$ and gray edges $\{a_i,a_{i+1}\}$ for $i\in [2n+1]$ and $2n+2\equiv 1$. Hence, $G$ is the superpositions of black and gray cycles, each of length $2n+1$.
The question is equivalent to finding a labeling of the vertices of $G$ with labels from $\{A,B\}$ (imposing a labeling of edges with $\{AA,AB,BB\}$) such that there are no two adjacent black edges both labeled $AA$, and there are no two adjacent gray edges both labeled $BB$ (call them *forbidden pair*). 

Let us show that the required labeling exists for any $n\geq 2$.

We construct graph $G'$ from $G$ by removing vertex $1$ with its incident edges, which we refer to as *special*. Then we remove black edges $\{3,4\}$, $\{5,6\}$, ..., $\{2n-1,2n\}$, and gray edges $\{a_{s+2},a_{s+3}\}$, $\{s_{s+4},a_{s+5}\}$, ..., $\{a_{s+2n-2},a_{s+2n-1}\}$, where $s:=a^{-1}(1)$, ie. $a_s=1$. We refer to these removed edges as *bridges* in $G$.

In the graph $G'$ every vertex is incident to one black and one gray edge, and so it is a disjoint union of alternating cycles of even length. We label vertices of each cycle alternatively with $A$'s and $B$'s starting at an arbitrary vertex, except for the cycle containing vertex $2$, which we label with $B$. Hence, all edges of $G'$ inherit label $AB$ from their endpoints. We propagate the labeling from $G'$ to $G$, additionally labeling vertex $1$ with $A$.

Each bridge in $G$ cannot be a part of forbidden pair, since both its adjacent edges of the same color have labels $AB$. On the other hand, special edges cannot form a forbidden pair either, since the black special edge $(1,2)$ has label $AB$, while the labels of gray special edges are either $AA$ and $AB$ (since vertex $1$ has label $A$). QED