By definition, for $SO(n)$ we have $g^T g=1$ for $g \in SO(n)$.
Given some vector $v \in V$ and some representation $R: SO(N) \rightarrow \mathrm{Lin}(V)$, the defining condition above tells us immediately
$$ v^T v \rightarrow (R(g)v)^T R(g)v = v^T v$$
$ \rightarrow $ The usual scalar product is invariant under $SO(n)$ transformations.
A useful way to think about representations is in terms of weights, where we use the eigenvectors of the Cartan generators of the corresponding algebra as basis for $V$. These basis vectors are labelled by the corresponding eigenvalues of the Cartan generators.
A trivial example is $SO(3)$, where we have for the 3-dimensional fundamental representation the basis vectors, $(1),(0),(-1)$, where I labelled them as explained above.
Therefore an arbitrary element of the 3-dim representation is
$$ v = \begin{pmatrix} a \\ b\\c \end{pmatrix} = a(1) + b(0) + c(-1) $$
Something invariant (belonging to the 1-dim representations)has always all labels zero, here $(0)$. The "naive" scalar product yields
$$ v^T v = a^2(1)^T(1) + b^2(0)^T(0) + c^2(-1)^T (-1) $$
I learned in this book that the labels simply get added if we multiply two vectors, therefore
$$ \rightarrow v^T v= a^2(2)+b^2(0)+c^2(-2) $$
This is certainly not something invariant.
As I wrote this, I noticed that the error must be that the label gets a minus sign if we transpose the state. This would yield
$$ \begin{align} v^T v & = a^2(1)^T(1) + b^2(0)^T(0) + c^2(-1)^T (-1) \\[8pt] & = a^2(-1)(1) + b^2(0)(0) + c^2(+1) (-1) = a^2(0) + b^2(0)+c^2(0) \end{align} $$
as it should be.
Therefore my question: Why does this sign change for the eigenvalues of the Cartan generators happen for the transposed vectors?