To find the ratio of the initial velocity to the initial acceleration of a particle moving along the $x$-axis, we need to differentiate the given position function with respect to time $t$.

Let's start by writing the position function:

$$x = \alpha t^4 + \beta t^2 + \gamma t + \delta$$

To find the velocity ($v$), we differentiate $x$ with respect to $t$:

$$v = \frac{dx}{dt} = \frac{d}{dt} (\alpha t^4 + \beta t^2 + \gamma t + \delta)$$

This gives us:

$$v = 4\alpha t^3 + 2\beta t + \gamma$$

The initial velocity is the velocity at $t = 0$:

$$v(0) = 4\alpha (0)^3 + 2\beta (0) + \gamma = \gamma$$

Next, to find the acceleration ($a$), we differentiate $v$ with respect to $t$:

$$a = \frac{dv}{dt} = \frac{d}{dt} (4\alpha t^3 + 2\beta t + \gamma)$$

This gives us:

$$a = 12\alpha t^2 + 2\beta$$

The initial acceleration is the acceleration at $t = 0$:

$$a(0) = 12\alpha (0)^2 + 2\beta = 2\beta$$

Now, we find the ratio of the initial velocity to the initial acceleration:

$$\text{Ratio} = \frac{v(0)}{a(0)} = \frac{\gamma}{2\beta}$$

Therefore, the correct answer is:

Option D: $\gamma: 2\beta$