 Solution

The Ricker population map two-cycle consists of the two stable points of the second iterate, exchanged by the original map. Thus generic positive orbits approach alternating valuesThere are exceptions: zero stays extinct, while an orbit starting at $\log 9$ or at one of its finite preimages eventually stays at that unstable fixed point.

To justify the large-time conclusion, put $b=9/e$ and $a=H(b)\simeq 1.087>1$. Every first image is at most $b$. A positive orbit below $a$ grows until it enters $[a,b]$; if it never entered, its monotone limit below $a<\log 9$ would contradict the fixed-point equation. The interval $[a,b]$ is invariant. On it $H$ is decreasing, so $H^2$ is increasing. Its only fixed points there are the three positive points already found; its graph signs make iterations below the unstable middle point converge to the lower stable point, and those above converge to the upper stable point. Interchanging even and odd subsequences under $H$ then gives the stated two-cycle. This is regular period-two behaviour, not chaos at this parameter.