So, by analogy the $I_0$ should exhibit the same behaviour as the amplitude in the force damped oscillation $\dots$
You are comparing two different things so let me explain.
For a mechanical system undergoing force oscillation the equation of motion is $m\ddot y+r\dot y+ky = f(t)$ and for a series LCR circuit the equation is $L\ddot q+R\dot q+\left(\frac 1C\right)q=v(t)$.
Thus you can see the similarity between the two systems with displacement $y$ being analogous to charge $q$, velocity $\dot y$ to current $i =\dot q$ and acceleration $\ddot y$ to rate of change of current $\dot i = \ddot q$.
And also $\rm mass \leftrightarrow inductance,\,friction\leftrightarrow resistance,\,spring\,constant \leftrightarrow 1/capacitance$.
Then you have a sinusoidal driving force $f(t)$ and a sinusoidal voltage source $v(t)$.
I will now consider the electrical system in steady state with $L=1\,{\rm H},\,C=1\,{\rm F}$ and $R=1\,\Omega$ for ease of analysis and using these values $x_0=\frac{1}{\sqrt{LC}}=1 \rm\,rad/s,$. Note the use of $x$ instead of $\omega$ for the frequency.
With the current $i$ as the reference for phase the phasor diagram looks like this:

The current in the circuit $i=\dfrac vZ= \dfrac{1}{\sqrt{1+\left(x-\frac 1x\right)^2}}$ if the applied voltage $v=1\,\rm V$.
The voltage across the resistor is $v_{\rm R}= i =\dfrac{1}{\sqrt{1+\left(x-\frac 1x\right)^2}}$,
The voltage across the capacitor is $v_{\rm C}=-\dfrac ix = -\dfrac{1}{x\sqrt{1+\left(x-\frac 1x\right)^2}}$ and this is also the charge stored on the capacitor.
The voltage across the inductor is $v_{\rm L}= ix = \dfrac{x}{\sqrt{1+\left(x-\frac 1x\right)^2}}$
$v_{\rm L}+v_{\rm C}= \dfrac{x}{\sqrt{1+\left(x-\frac 1x\right)^2}} -\dfrac{1}{x\sqrt{1+\left(x-\frac 1x\right)^2}} $
Power $= i^2 = \dfrac{1}{1+\left(x-\frac 1x\right)^2}$.
This is what the graphs as the frequency $x$ is varied look like.

You see that current (velocity in the mechanical case) resonance and power resonance occur at $x_0=\frac{1}{\sqrt{LC}}=1$ but charge (displacement in the mechanical case) resonance occurs at a lower frequency.
I think that a lot of the confusion arises because it is relatively easy to measure currents in electrical circuits but much more difficult to measure velocities in mechanical cases and so in mechanical examples the amplitude (maximum displacement) is measured.
You quoted the equation for amplitude as
$A=\dfrac{F_0}{m\sqrt{(\omega_0^2-\omega^2)^2+(\omega\gamma)^2}}=\dfrac{1}{\sqrt{(1-\omega^2)^2+\omega^2}}=\dfrac{1}{\omega\sqrt{(
\frac 1 \omega -\omega)^2 +1}} $ after putting all the constants equal to one.
Note that this is the displacement not velocity formula.