Can I use $Q^T = Q^{-1}$ ($Q$ is orthogonal eigenvector matrix) to prove a certain implication?

Question

Is the chain of implications below correct or does it break down somewhere along the way?

Let $A$ be a square matrix, $D$ a diagonal matrix and $P$ an eigenvector matrix. Consider $A = P^{-1}DP$ which implies $A = PDP^{-1}$ and so $A = P^{-1}DP = PDP^{-1}.$ Then $A = Q^{-1}DQ = QDQ^{-1}$ which implies $A = Q^TDQ = QDQ^T$ since $Q^T = Q^{-1}.$

Answer 1

One way to see that your reasoning is wrong is that, if it were true, then every diagonalizable matrix would be symmetric (because if $D$ is diagonal then $D=D^T$), which is clearly wrong.

Answer 2

Your chain of reasoning breaks down right away. $A=P^{-1}DP$ does not imply $A=PDP^{-1}$. You make a similar error in one of your earlier questions.