i don't know the exact solution but i think the idea is to put numbers around a circle then counting the cycles

Let $gcd(n,k)=d$.The answer is all $(n,k)$ with $2d\nmid n$. $\textbf{\underline{Necessity}}$ Suppose $2d\mid n$.Consider the following 2 permutitions: $$(\textcolor{red}{1,2,\dots d},\textcolor{blue}{d+1,d+2,\dots 2d},\textcolor{red}{2d+1,2d+2,\dots 3d},\textcolor{blue}{3d+1,3d+2,\dots 4d},\textcolor{red}{4d+1},\dots\ ,\dots \textcolor{blue}{n})$$ $$(\textcolor{blue}{d+1,d+2,\dots 2d},\textcolor{red}{1,2,\dots d},\textcolor{blue}{3d+1,3d+2,\dots 4d},\textcolor{red}{2d+1,2d+2,\dots 3d},\textcolor{blue}{5d+1},\dots\ ,\dots \textcolor{red}{n-d})$$ Where in the first permutition we colour $kd+1,\dots (k+1)d$ red when $k$ is even and blue otherwise.In the second permutition we flip the 1st red and 1st blue block; 2nd red and 2nd blue block,etc.As $\frac{n}{d}$ even hence, $n$ must be blue.It is easy to verify that this 2 permutition gives same $k$-mutation.$\blacksquare$ $\textbf{\underline{Sufficiency}}$ Let $2d\nmid n$.So $\frac{n}{d}$=odd. FTSOC assume for 2 different permutition $\mathfrak A$ and $\mathfrak B$ let: $$a_i+a_{i+k}=b_i+b_{i+k}\ \ \ \forall i=1,2\dots n$$As this 2 permutition are different for some $r$ less than $n$ we have: $$a_r>b_r$$Then we must have :\begin{align*}a_{r+k}&<b_{r+k}\\ a_{r+2k}&>b_{r+2k}\\ a_{r+3k}&<b_{r+3k}\\ \dots\\ a_{r+\frac{n}{d}\times k}&<b_{r+\frac{n}{d}\times k} \end{align*}The last inequality holds since $\frac{n}{d}$ is odd.The indices are taken $\mod{n}$. But this is a contradiction as we assumed $a_r>b_r$. $\blacksquare$

Solution. The answer is all $(n,k)$ such that $\frac{n}{(n,k)}$ is odd. Claim 1. If $\frac{n}{(n,k)}$ is odd, then $(n,k)$ works. Proof. For the sake of contradiction, let $(a_1,a_2,\dots,a_n)$ and $(b_1,b_2,\dots,b_n)$ have the same $k$-mutation and assume we also have that $\frac{n}{(n,k)}$ is odd. WLOG $a_1 > b_1$. Note that since they have the same $k$-mutation, we have $a_{1+xk} > b_{1+xk}$ if and only if $x$ is even. Let $\ell >0$ be the smallest such that $1+\ell k \equiv 1 \pmod{n}$ (it's not hard to see that such $\ell$ must exist). Then $$\ell k \equiv 0 \pmod{n}\iff \ell \equiv 0 \pmod{\frac{n}{(n,k)}}$$The smallest $\ell$ is $\frac{n}{(n,k)}$. If $\ell$ is odd, then it's a contradiction. Claim 2. $(2x,1)$ doesn't work for all $x \in \mathbb{N}$. Proof. Take the two permutations: $$(1,2,3,4,5,6,7,8,9,10,\dots, 2x)$$$$(2,1,4,3,6,5,8,7,10,9,\dots,2x,2x-1)$$which have the same $1$-mutations. Claim 3. If $\frac{n}{(n,k)}$ is even, then $(n,k)$ doesn't work. Proof. It's not hard to see that we can "decompose" into Claim 2., hence it doesn't work.