Proving that if $\hat\pi=\bar\alpha$, then $\pi=id$ and $(\alpha_1,\alpha_2,...,\alpha_n)=(id_F,id_F,..,id_F)$

Question

I'm having some trouble with the following problem:

Let $F$ be a finite field and consider the set $F^n$. Let $\pi\in S_n$ be a permutation, and define the map $$\hat\pi:F^n\to F^n$$ such that $\hat\pi(x_1,x_2,...,x_n)=(x_{\pi(1)},x_{\pi(2)},...,x_{\pi(n)})$.

Now let $\alpha_1,...,\alpha_n:F\to F$ be bijections, and define the map $$\bar\alpha:F^n\to F^n$$ such that $\bar\alpha(x_1,x_2,...,x_n)=(\alpha_1(x_1),\alpha_2(x_2),...,\alpha_n(x_n))$.

Prove that, if $\hat\pi=\bar\alpha$, then $\pi=id$ and $(\alpha_1,\alpha_2,...,\alpha_n)=(id_F,id_F,..,id_F)$.

Since $\hat\pi=\bar\alpha$ are the same for all inputs $x\in F^n$, i think that maybe by choosing some cleaver inputs for the functions , we can show that they must be the identity, but I'm not sure. How can this be done?

Answer 1

Taking $x_1 = x_2 = \cdots = x_n = x$, you get $$ \hat\pi(x,x,\ldots,x)=(x,x,...,x) \quad \text{and} \quad \bar\alpha(x,x,\ldots ,x) = (\alpha_1(x),\alpha_2(x),...,\alpha_n(x)) $$ Thus if if $\hat\pi=\bar\alpha$, one has $(\alpha_1,\alpha_2,...,\alpha_n)=(id_F,id_F,..,id_F)$. It follows that $$ (x_{\pi(1)},x_{\pi(2)},\ldots,x_{\pi(n)}) = \hat\pi(x_1,x_2,\ldots,x_n) = \bar\alpha(x_1, x_2,\ldots, x_n)=(x_1,x_2,\ldots,x_n) $$ Consequently, $\pi=id$.

Note that $F$ could be any finite set. The condition that $F$ is a field is useless.