= Solution
For a <normalization>[normalized] state the position <probability density> is $\boxed{f=|\psi|^2}$. With a real potential, the <Time-dependent Schrödinger equation> and its conjugate give
$$
f_t=\frac{i\hbar}{2m}(\overline\psi\psi_{xx}-\psi\overline{\psi_{xx}})=-\partial_x\left[\frac{\hbar}{2mi}(\overline\psi\psi_x-\psi\overline{\psi_x})\right].
$$
This is the <continuity equation> $\boxed{f_t+j_x=0}$. Integrating over the line yields $d\|\psi\|^2/dt=-j(\infty)+j(-\infty)=0$ under the stated decay. A real-valued <wave function> gives $\overline\psi\psi_x-\psi\overline{\psi_x}=0$, so its instantaneous current vanishes; the same is true up to a spatially constant <phase>.
Put $\alpha=m\omega/\hbar$. <Complete> the square in the overlap:
$$
\overline{G(u,x)}G(s,x)=|N|^2\overline{\sqrt u}\sqrt s\,e^{-\alpha[(x+\overline u+s)^2-2\overline u s]}.
$$
The <Gaussian integral> is $\sqrt{\pi/\alpha}$, also for complex displacement by <analytic continuation>. Hence
$$
\boxed{\langle G(u),G(s)\rangle=|N|^2\overline{\sqrt u}\sqrt s\sqrt{\pi/\alpha}\,e^{2\alpha\overline u s}.}
$$
The product of square roots fixes the branch convention implicit in the printed square root of $\overline u s$. At $u=s\ne0$ the <norm> squared is $|N|^2|s|\sqrt{\pi/\alpha}\,e^{2\alpha|s|^2}$, which becomes one for the stated $N$. In particular
$$
\boxed{|G(s,x)|^2=\sqrt{\alpha/\pi}\,e^{-\alpha(x+2\operatorname{Re}s)^2}.}
$$
The centre is $-2\operatorname{Re}s$ and its position <variance> is $1/(2\alpha)$.
When differentiating in $s$, keep $N$ fixed on the chosen constant-radius orbit. Then
$$
G_x=-\alpha(x+2s)G,\quad G_{xx}=[\alpha^2(x+2s)^2-\alpha]G,\quad sG_s=(\tfrac12-2\alpha sx-2\alpha s^2)G.
$$
Substitution gives $HG=\hbar\omega sG_s$. Since $\dot s=-i\omega s$, the <chain rule> gives $\boxed{i\hbar G_t=HG}$. Continue $\sqrt{s(t)}$ continuously along the time orbit; its sign after a full orbit supplies the usual oscillator zero-point <phase>. The case of zero displacement is obtained as a limit, since the separate printed $N$ and $\sqrt s$ expressions are singular at $s=0$.
Finally $j=(\hbar/m)\operatorname{Im}(\overline G G_x)=-2\omega\operatorname{Im}s\,|G|^2$. For real nonzero $a$ this is
$$
\boxed{j(x,t)=2a\omega\sin(\omega t)\sqrt{\frac{m\omega}{\pi\hbar}}\exp\!\left[-\frac{m\omega}{\hbar}(x+2a\cos\omega t)^2\right].}
$$
\b[The printed final current omits the Gaussian <normalization> factor] $\sqrt{m\omega/(\pi\hbar)}$. Retaining that factor is required both by the <normalization>[normalized] state and by the <continuity equation>. This current is the <normal distribution>[Gaussian probability density] times the velocity of its moving centre, $2a\omega\sin\omega t$.
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