We are given a permutation group G acting on a finite set X. Finite set $\Omega$ contains all mappings $\omega$ from the set X to finite set K. We define mapping $\hat{g}$ on the set $\Omega$ in the following way

$(\hat{g}(\omega))(x)= \omega(g(x))$

Question: Is the statement below true?

If a,b are in G then $((\hat{a}\hat{b})(\omega))(x)=((\widehat{ab}(\omega))(x)$

I will show that it is not true:

$((\widehat{ab}(\omega))(x)=\omega((ab)(x))=\omega(a(b(x)))$

$((\hat{a}\hat{b})(\omega))(x)=(\hat{a}(\hat{b}(\omega)))(x)=(\hat{a}(\omega_p))(x)=\omega_p(a(x))=\omega(b(a(x)))$

$\hat{b}(\omega)=\omega_p$ and $(\hat{b}(\omega))(x)=\omega_p(x)=\omega(b(x))$

Are my deductions correct? If they are correct then mapping which is defined above cannot be a ~~homeomorphism~~ isomorphism?