Transfer Game - version 2

Let initially A had $\$x$ and B had $\$(360-x)$

After A gave the money to B, he had $\$0.8x$ and B had $\$(360-0.8x)$

After B gave money to A, the later (A) has $\$(0.64x+72)$

So, according to the given condition,

$x=0.64x+72\implies 0.36x=72\implies x=\dfrac {72}{0.36}=\boxed {200}$ .

Let the initial and end amounts $A$ and $B$ have be $a_0$ and $b_0$ , and $a_1$ and $b_1$ . Then $a_0+b_0 = 360$ . Then

$\begin{cases} b_1 = 0.8(0.2a_0+b_0) = 0.16 a_0 + 0.8 b_0 \\ a_1 = 0.8a_0 + 0.2(0.2a_0+b_0) = 0.84a+0.2b_0 \end{cases}$

Since $\begin{cases} b_1 = b_0 & \implies 0.16 a_0 + 0.8 b_0 = b_0 & \implies b_0 = 0.8 a_0 \\ a_1 = a_0 & \implies 0.84a+0.2b_0 = a_0 & \implies b_0 = 0.8 a_0 \end{cases}$

Therefore, $a_0+b_0 = 1.8 a_0 = 360 \implies a = \boxed{200}$