Let $\Phi$ be a matrix n-dimensional representation of the group G. We construct a representation $\Psi$ of $G$ on the space of square matrices of order n, such that $$\Psi_g(A)=\Phi(g)A^t\Phi(g) \text{ as } A\in M_n(K) $$ express $\chi_\Psi$ through $\chi_\Phi$
This is my attempt so far. $$\chi_\Psi (g)=\chi (\Phi(g)A^t\Phi(g))=\chi (\Phi^2(g)A^t).$$ I know that $\chi(A^t)=\chi(A^t)$ At this point I am now stuck. Is answer $\chi_\Psi=\chi_\Phi^2?$ Help me please.
Sorry for the poor english and if I wasn't clear, please tell me.
Edit: incidentally, the way you seem to have interpreted the definition, it doesn't even define a left action, since you don't have $\Psi_{gh}=\Psi_g\Psi_h$. I am sure the intention was for the transpose sign to belong to the second $\Psi(g)$, rather than to $A$.
The character of a representation is the trace of the matrix with respect to any basis on your vector space. The vector space attached to $\Psi$ is $n^2$-dimensional. So any group element is represented by an $n^2$-by-$n^2$ matrix, and the character is a sum of $n^2$ values. I don't understand your calculation, but I am pretty sure that that's not what you were doing.
So question number 1 is: can you give yourself a nice basis on the space of $n\times n$ matrices? After that you have to act by $g$ on each such basis element, and then express the result again in terms of your basis and take the coefficient of the basis element that you have acted on. That's what it means to read off a diagonal value of $\Psi_g$.
Your calculation will be simplified if you assume that $\Phi(g)$ is diagonal. You can always do that for any given $g$ by choosing a nice basis on your original vector space (of course you may not be able to diagonalise all $g\in G$ simultaneously, but that's ok).