Observe first that $G A \perp G C$ if and only if $5 A C^{2} = A B^{2} + B C^{2}$. Indeed,

$$G A \perp G C \Leftrightarrow \frac{4}{9} m_{a}^{2} + \frac{4}{9} m_{c}^{2} = b^{2} \Leftrightarrow 5 b^{2} = a^{2} + c^{2}$$

Moreover,

$$G B^{2} = \frac{4}{9} m_{b}^{2} = \frac{2 a^{2} + 2 c^{2} - b^{2}}{9} = \frac{9 b^{2}}{9} = b^{2}$$

hence $G B = A C = C A^{\prime}$ (1). Let $C^{\prime}$ be the intersection point of the lines $G C$ and $A B$. Then $C C^{\prime}$ is the middle line of the triangle $A B A^{\prime}$, hence $G C \parallel B A^{\prime}$. Consequently, $G C A^{\prime} B$ is a trapezoid. From (1) we find that $G C A^{\prime} B$ is isosceles, thus cyclic, as needed.

Conversely, since $G C A^{\prime} B$ is a cyclic trapezoid, then it is also isosceles. Thus $C A^{\prime} = G B$, which leads to (1).

Comment: An alternate proof is as follows:

Let $M$ be the midpoint of $A C$. Then the triangles $M C G$ and $M A^{\prime} B$ are similar. So $G C$ is parallel to $A^{\prime} B$.

$G A \perp G C$ if and only if $G M = M C$. By the above similarity, this happens if and only if $A^{\prime} C = G B$; if and only if the trapezoid is cyclic.