Nice problem We will build a series of swaps out of the ones we are given. Let $a= \text{max} (k,n), b =\min (k,n)$. Consider two numbers which differ by $a-b$: Let them be $x, x+a-b$. It's clear that one of $x-b, x+a$ is in the range $[1, k+n]$, WLOG suppose $x-b$ is in the range. We can switch the pair $(x,x-b)$, then switch the pair $(x-b,x+a-b)$, and then switch the positions of $(x,x-b)$ again. The end result of this sequence of moves is that the pair $(x,x+a-b)$ has switched positions. Therefore if we can swap numbers differing by $a$ and $b$, we can swap numbers differing by $a-b$. Now it's clear what we should do: We can replace $(a,b)$ with a new tuple $(a', b') = (\text{max} (b,a-b), \min (b,a-b))$. It's clear that $(a',b')$ has the same property as $(a,b)$: we can swap any numbers differing by $a'$ or $b'$. But as we keep replacing the tuple, $(a,b)$ is following precisely the Euclidean Algorithm; therefore, eventually the process terminates when $a=b=1$, since we assumed $\gcd (k,n)=1$. At this point, we know that we can build a series of moves which swaps any two numbers differing by $1$ but does not change the positions of other numbers. Then it's easy to place the numbers in increasing order with the following algorithm: Suppose the first $m$ numbers are now in increasing order for $m\ge 0$. Let the $m+1$th number be $x$, so clearly $x>m+1$. We can swap the following pairs: $(x, x-1), (x-1,x-2), ... (m+2,m+1)$. When we perform this switch, the number $m+1$ will go to the original position of $x$, which is position $m+1$. Therefore, we have now sorted the first $m+1$ numbers in increasing order. We can repeat the process until $m=k+n$, at which point the whole list is sorted.

Another nice solution: Consider a graph with vertices $V_i, 1\le i\le n + k$, and connect two of them if and only if their indices differ by $n$ or $k$. We can easily see it is connected for the following reason: All vertices connect back to one of $V_1, V_2, \ldots, V_n$. Now, $\{1, 1 + k, 1 + 2k, \ldots, 1 + (n - 1)k\}$ forms a complete residue system modulo $k$, and hence it is easy to see that by shifting from $i$ up by $k$ multiple times (and shifting backwards by $n$ whenever we hit an integer from $[n + 1, n + k]$) we hit all of $V_1, \ldots, V_n$ in our path. So everything is connected. Now take a spanning tree (i.e. delete one edge from the lone cycle that exists). Then, at every turn, we match up a leaf up with its number by proper swaps and then delete it. This works, and is more general than above (although Euclid's Lemma is nice).

By Bezout, we have that there exist positive integers $x,y$ such that $kx-ny=1$ or $ny-kx=1$. Suppose wlog we are in the first case, i.e. $kx-ny=1$. Construct the following sequence: $x_1=1=kx-ny$ and if $x_n=k\alpha-n\beta$, then $x_{n+1} \in \{ k(\alpha+1)-n\beta,\ k\alpha-n (\beta+1)\}$ with $1\le x_{n+1}\le k+n$. It is easy to see that $x_{n+1}$ is thus uniquely defined. Note that if $x_i=ks-nt$, then $s+t=x+y+i-1$. I claim that for $i\le k+n,\ x_i\ne x_j$ for any $j<i$. If $x_i=ka-nb=kc-nd=x_j$, from $(k,n)$ we get that $a=c+n\alpha$ and $b=d+k\alpha$, whence $x+y+i-1=a+b\ge c+d+k+n=x+y+j-1+k+n$, contradiction. Thereby, $x_1,...,x_{k+n}$ is a permutation of $1,2,...,k+n$ in which $x_{i+1}$ can commute with $x_i$ and $x_{i-1}$. The numbers are initially in some order $x_{\sigma(1)},...\ ,x_{\sigma(k+n)}$ and we wish to permute them in an order $x_{\tau(1)},...\ ,x_{\tau(k+n)}$ such that $x_{\tau(i)}=i$. One can prove that $(S_m,\circ)$ is generated by the set of transpositions $\{ (1\ 2),(2\ 3),...,(m-1\ m)\}$ by using that any permutation can be written as the product of some transpositions,$(k\ l)=(1 k) (1\ l)(1\ k)$ combined with $(1\ k+1)=(1\ k)(k\ k+1)(1\ k)$ and induction. Thus, we can apply some transpositions in order to obtain $\sigma^{-1} \tau$, which means that we can get $\tau$ from $\sigma$ after some switches.