For $SU(2)$, the fundamental representation is a quaternionic representation. Which means there is a preserved skew symmetric form, written as a matrix:
$$ \varepsilon = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right) $$ i.e. for any element $U \in SU(2)$ $$ \quad U^T \varepsilon \>U = \varepsilon $$
However there is simultaneously a symmetric invariant bi-linear form:
$$ g = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right) $$ i.e. for any element $U \in SU(2)$ $$ \quad U^\dagger g \>U = g $$
How can there be simultaneously two invariant bi-linear forms given the well known theorems regarding the Frobenius-Schur indicator?
I've realized this question only arises from forgetting a minor detail: the answer to this is that the $g$ is not bilinear, it is sesquilinear, for which the Frobenius-Schur indicator tells us nothing about. Thus having both is fine.