Let $ V $ be an finite-dimensional inner product space over $ \mathbb{C} $.
Let $ T : V \to V $ be a linear transformation and assume that for every $ v $, $ \lVert Tv \rVert =\lVert T^*v\rVert$. I want to prove that $ T $ is a normal transformation.
My effort:
Suppose, to the contrary, that $ T $ is not normal and therefore there exists a $ \vec{u} $ such that $ TT^*(\vec{u}) \neq T^*T(\vec{u}) $.
$ \lVert Tv\rVert^2= \langle Tv, Tv \rangle = \langle v, T^*Tv\rangle$
$ \lVert T^*v \rVert^2=\langle T^*v, T^*v\rangle=\langle v, TT^*v\rangle$
which means $ \lVert Tv\rVert^2 - \lVert T^*v \rVert^2==\langle v, (T^*Tv -TT^*v) \rangle$
I now want to show that $ T^*Tv- TT^*v \not \perp u $, but I don't really know how? Could you give me a $ \textbf{hint} $?
From the last equation you have derived you get that the RHS inner product is zero for all $v\in V$.
The hint is the term "Polarisation" to finish your proof,
cf wikipedia or math SE's "Polarization identity in Hilbert space" if you'd want to stay on this site.
Furthermore, the post If $(Tx \mathbin{|} x) = 0$ for all $x$ then $T = 0$ is a seamless sequel of your post.