= Solution
Put $h(r)=1-\mu/r^2$. For the equatorial <geodesic>, the <geodesic> <Lagrangian> is $\mathcal L=\tfrac12[-h\dot t^2+h^{-1}\dot r^2+r^2\dot\phi^2]$, where dots denote <derivatives> with respect to the <affine parameter>. The cyclic coordinates $t,\phi$ give constants $p_t=-h\dot t=-E$ and $p_\phi=r^2\dot\phi=L$. A third constant is the tangent <norm> $g_{ab}\dot x^a\dot x^b=-\kappa$, preserved by the <geodesic> equation and <metric compatibility>. For a timelike <geodesic> rescale the <affine parameter> to <proper time> so $\kappa=1$; for a null <geodesic> $\kappa=0$.
Substitute $\dot t=E/h$, $\dot\phi=L/r^2$ into the <norm> identity and multiply by $h$. It gives the <effective potential> equation
$$
\boxed{\frac12\dot r^2+V(r)=\frac12E^2,\qquad
V(r)=\frac12\left(1-\frac\mu{r^2}\right)\left(\kappa+\frac{L^2}{r^2}\right)}.
$$
Outside the horizon $r=\sqrt\mu$, its <derivative> is $V'=-(L^2-\kappa\mu)/r^3+2\mu L^2/r^5$. A nontrivial circular null <geodesic> requires $L\ne0$, giving
$$
\boxed{r_{\rm null}=\sqrt{2\mu},\qquad V''(r_{\rm null})=-\frac{L^2}{2\mu^2}<0}.
$$
For a circular timelike <geodesic>,
$$
\boxed{r_{\rm time}^2=\frac{2\mu L^2}{L^2-\mu},\qquad L^2>\mu}.
$$
At it $V''=-4\mu L^2/r^6<0$, and its radius exceeds $\sqrt{2\mu}$. If $L^2\le\mu$ no finite circular timelike radius exists. Thus \b[all finite circular null and timelike orbits in this five-dimensional exterior are radially unstable]. The <energy> of a circular orbit is fixed by $E^2=2V(r)$, not independently arbitrary.
For comparison, with four-dimensional Schwarzschild horizon radius $r_s$, the photon circle is at $3r_s/2$ and is unstable. Circular timelike orbits exist for $r>3r_s/2$; they are unstable below $3r_s$, marginal at $3r_s$ and stable above $3r_s$. The five-dimensional potential has no corresponding stable outer circular timelike branch.
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