I'm stuck with this question. I do believe the claim is correct but I don't know how to show it.
If we know that $T^2=Id$, then $T=T^{-1}$on the one hand, and on the other hand we know that $T^2$ is orthogonal, hence $(T^2)^*=(T^2)^{-1}$ which implies that $(T^{-1})^2=(T^*)^2$. I can't find the way to take it to the next step and show that $T^{-1}=T^*$ to get that $T$ is also orthogonal, nor am I able to conclude anything about $T, T^*$'s commutativity.
No, one can find counterexamples in 2x2 matrices.
Take \begin{pmatrix} 1 & -2\cr 0 & -1 \end{pmatrix} This is the symmetry w.r.t the line $y=x$ (anti-invariant) across $x=0$ (invariant). It sends $I\to I;\ I+J\to -(I+J)$.
One can prove that ($T^2=I$ and $T$ normal) implies ($T$ is an orthogonal symmetry). Then if you take any symmetry which is not orthogonal, you get a counterexample.