Perhaps the title is slightly odd, but I couldn't think of a better one. Suppose we have a total recursive function $f:N^2 \rightarrow N$. Define the n-th row of $f$ as the function $F_n:N \rightarrow N$ where:
$$F_n(x)=f(x,n) \qquad for \; all \; x$$
Now suppose that every single row of F is unique. What that means precisely is that for any two distinct row numbers $a$ and $b$ (where $a\neq b$) we must have $F_a \neq F_b$.
Now consider an arbitrary total recursive function $g:N^2 \rightarrow N$ (not given in advance). It is given that $g$ "shuffles" the rows of $f$ without changing any of them. Therefore every single row of the function $g$ is also unique (and the set of rows that occur in $f$ and $g$ must be the same).
Now given this there must be a unique permutation function $P:N \rightarrow N$ such that when it is applied to a given row number of $f$, gives us the position of occurrence of same row in $g$. For example if some function $F_0:N \rightarrow N$ occurred as $0$-th row in $f$, and the same function occurred as $20$-th row in $g$, we will have $P(0)=20$.If some function $F_1:N \rightarrow N$ occurred as $1$st row in $f$, and the same function occurred as $0$-th row in $g$, we will have $P(1)=0$. And so on.
For any given $f$ and $g$ that satisfy the conditions mentioned above, the question asks whether $P$ is always recursive or not? If it is, then it has to be show why. If it isn't at least one example has to be given by specifying the functions $f$ and $g$.
We are given total recursive functions $f:\mathbb{N}^2 \rightarrow \mathbb{N}$ and $g: \mathbb{N}^2 \rightarrow \mathbb{N}$ such that (1) every row $f_n = \lambda x. f(x,n)$ is distinct, (2) the rows $g_n$ of $g$ are exactly equal to the rows of $f$, but maybe in a different order, and consequently (3) there is a unique permutation $P$ of $\mathbb{N}$ such that $f_n = g_{P(n)}$ for all $n$. Your question is whether $P$ is recursive. The answer is no, and the general reason is that although we will be able to recognize, sooner or later, that a given $f_n$ and $g_m$ are $\it not$ equal, we might never be able to recognize when they $\it are$.
The problem of finding $P$ is, in the worst case, at the level of Halting Problem, which we denote by $K$. You would prove this by showing (on the one hand) that $P$ can be computed from the $K$-computable set $A = \{\langle m,n\rangle \in \mathbb{N}^2 : \forall $x$. f_n(x) = g_m(x)\}$, and (on the other) constructing a pair $f,g$ such that if $n \in K$, i.e., if the $n$-th Turing machine halts, then something like $P(n)$ or $P(2n)$ is greater than the machine's runtime. In this case we could use $P$ to decide whether a given $n$ is in $K$ by running the $n$-th Turing machine for something like $P(n)$ or $P(2n)$ steps and checking whether it has halted yet.