What would you like to understand?

If $W$ is a $G$-invariant subspace and the inner product is $G$-invariant, then for any $y \in W^{\perp}$ and $g \in G$, the vector $\rho(g)y$ remains orthogonal to every $w \in W$. This is shown by rewriting $(\rho(g)y, w)$ as $(y, \rho(g^{-1})w)$.

Conditions: $W \subseteq V$ is a $G$-invariant subspace ($\rho(g)W \subseteq W$ for all $g \in G$).; The inner product $(\cdot, \cdot)$ on $V$ is $G$-invariant ($(\rho(g)v, \rho(g)w) = (v, w)$).; $W^{\perp} = \{y \in V \mid (y, w) = 0 \text{ for all } w \in W\}$.

If $W$ is a $G$-invariant subspace and the inner product is $G$-invariant, then for any $y \in W^{\perp}$ and $g \in G$, the vector $\rho(g)y$ remains orthogonal to every $w \in W$. This is shown by rewriting $(\rho(g)y, w)$ as $(y, \rho(g^{-1})w)$.

Conditions: $W \subseteq V$ is a $G$-invariant subspace ($\rho(g)W \subseteq W$ for all $g \in G$).; The inner product $(\cdot, \cdot)$ on $V$ is $G$-invariant ($(\rho(g)v, \rho(g)w) = (v, w)$).; $W^{\perp} = \{y \in V \mid (y, w) = 0 \text{ for all } w \in W\}$.