Suppose $T_1, T_2$ are normal operators on $\mathcal L(\mathbb{F}^3)$ and both operators have $2, 5, 7$ as eigenvalues. Prove that there exists an isometry $S\in \mathcal L(\mathbb{F}^3)$ such that $T_1 = S^*T_2S$
I don't know how to approach this problem and Question: If $T_1, T_2$ have the same eigenvalues, does it mean their matrix representation is the same w.r.t the same orthonormal basis of $\mathbb{F}^3$.
If the matrix representation of two operators with respect to some basis is the same, then the operators are equal and this doesn't have to be the case, even in your situation.
In your case, call the distinct eigenvalues $\lambda_1,\lambda_2,\lambda_3$. Since $T_1$ has distinct eigenvalues, it is diagonalizable. Since $T_1$ is normal, you can choose an orthonormal basis $(v_1,v_2,v_3)$ of $\mathbb{F}^3$ consisting of eigenvectors of $T_1$ (with $T_1v_i = \lambda_i v_i$). The same argument applies to $T_2$ resulting in an orthonormal basis $(w_1,w_2,w_3)$ of eigenvectors of $T_2$ (with $T_2 w_i = \lambda_i w_i$). Define an operator $S \colon \mathbb{F}^3 \rightarrow \mathbb{F}^3$ by $S(v_i) = w_i$. Since $S$ sends an orthonormal basis to an orthonormal basis, the map $S$ is an isometry and we have
$$ (S^{*}T_2S)(v_i) = S^{*}T(w_i) = S^{*}(\lambda_i w_i) = \lambda_i S^{-1}(w_i) = \lambda_i v_i = T_1(v_i) $$
and so $T_1 = S^{*}T_2S$.