Define an isomorphism between the complete bipartite graph $K_{n,n}$ and the Circulant graph with $2n$ vertices $[0]$ to $[2n-1]$ and jumps in the set $S := \{2k+1 : 0\leq k\leq \lfloor\frac{n-1}2\rfloor\},$ denoted $C_{2n}(S).$ Assume that $2n-1$ for simplicity (so that $C_{2n}(S)$ indeed has distinct vertices).
We define a bijection $f : V(K_{n,n}) \to V(C_{2n}(S))$ as follows. Let $(A,B)$ be a bipartition of $K_{n,n}.$ Let $x\in V(K_{n,n}).$ Enumerate the $n$ vertices in $A$ and $B$ as $\{a_1,\cdots, a_n\}$ and $\{b_1,\cdots, b_n\}$ respectively. Let $x \in V(K_{n,n}).$ Then 1) $x \in A$ or 2) $x \in B.$ If $x \in A,$ then $x = a_i$ for some $i$ by the definition of the $a_i$'s. We map $x$ to $[2(i-1)+1]$, which is clearly in the vertex set as $1\leq i\leq n.$ If $x \in B, $ then we may find $1\leq j\leq n$ so that $x=b_j$ and we can map $x$ to $[2j].$ And for any $v \in V(C_{2n}(S)),$ if $v = 2k+1$ for some $k$, map $v$ to $a_{k+1}\in A$ and if $v = 2k$ for some $k$, map $v$ to $b_k.$
We now claim that $\{v,w\}$ is an edge in $E(K_{n,n})$ iff $\{f(v),f(w)\}$ is an edge in $E(C_{2n}(S)).$ Suppose $\{v,w\}$ is an edge in $E(K_{n,n}).$ Then WLOG we may assume (by switching $v$ and $w$ if necessary) $v \in A$ and $w \in B.$ Then $v = a_i$ for some $i$ and $w = b_j$ for some $j$. We have by the definition of $f$ that $f(v) = 2(i-1)+1$ and $f(w) = 2j.$ Indeed, these are adjacent because $[2i-1 + (2(j-i)+1)] = [2j]$ and $[2(j-i)+1]\in S$ so by the definition of a circulant graph, $f(w)$ and $f(v)$ are adjacent.
Similarly, if $\{f(v), f(w)\}\in E(C_{2n}(S)),$ then we may assume WLOG that $f(v) = [2k+1]$ and $f(w) = [2j]$ for some $0\leq k\leq \lfloor\frac{n-1}2\rfloor$ and $1\leq j\leq n$. Then by the definition of $f, v \in A$ and $w \in B.$
Is this solution correct? Note that we could have defined $f(b_j) = 2(j-1)$, but since $[2n] = [0],$ it doesn't really matter.