Can different control points lead to the same Bézier curve?

The equation for the cubic Bézier curve is $$ F(u) = (1-u)^3 P_0 + 3 (1-u)^2 u P_1 + 3 (1-u)u^2 P_2 + u^3 P_3 $$ and therefore $$F(0) = P_0, \quad F(1) = P_3$$ and, by differentiation, $$ F'(0) = 3(P_1 - P_0), \quad F'(1) = 3(P_3 - P_2) $$

Therefore all control points are uniquely determined by the function $F$.

But, as you already noted, there can be two different Bézier curves having the same image, e.g. if all control points are on a line. I don't have an answer for the question

When are the control points uniquely determined by the image of a Bézier curve?

Suppose $\mathbf{P}_0, \ldots \mathbf{P}_m$, and $\mathbf{Q}_0 \ldots \mathbf{Q}_m$ are two sets of control points that produce the same Bézier curve of degree $m$, in the sense that $$ \sum_{i=0}^m b^m_i(t)\mathbf{P}_i = \sum_{i=0}^m b^m_i(t)\mathbf{Q}_i \quad \text{for all $t \in[0,1]$} $$ Then we have $$ \sum_{i=0}^m b^m_i(t)(\mathbf{P}_i - \mathbf{Q}_i) = \mathbf{0} \quad \text{for all $t \in[0,1]$} $$ This implies that $\mathbf{P}_i = \mathbf{Q}_i$ for $i=0,1, \ldots, m$ since the Bernstein polynomials $b^m_0, \ldots b^m_m$ are linearly independent.

There are cases where two different sets of control points will produce the same trace/image/trajectory. Take a given curve, and compose it with two different polynomials that both map $[0,1]$ to itself. Then clearly you'll get the same image, but the parametric equations will be different, so the control points will also be different.

A specific example: consider the two degree 4 curves whose control points are

$(0,0)$, $(1,0)$, $(\tfrac53,\tfrac23)$, $(2,1)$, $(2,1)$

$(0,0)$, $(0,0)$, $(\tfrac13,0)$, $(1,0)$, $(2,1)$.

Routine calculations show that these two curves both represent the portion of the parabola $y = \tfrac14 x^2$ that lies between the points $(0,0)$ and $(2,1)$. The first one has equation $G(t) = (4t - 2t^2, 4t^2 - 4t^3 + t^4)$ and the second one has equation $H(t) = (2t^2, t^4)$.

These curves are formed by two different reparameterizations of the basic curve $F(t) = (2t, t^2)$: $$ F(t) = (2t, t^2) \;,\; t = 2u - u^2 \;\; \Rightarrow \;\; G(u) = (4u - 2u^2, 4u^2 - 4u^3 + u^4) $$ $$ F(t) = (2t, t^2) \;,\; t = v^2 \;\; \Rightarrow \;\; H(v) = (2v^2, v^4) $$ This is a form of degree elevation (though not the usual form). The degree of the final curve is the product of the degree of the original one and the degree of the reparametrization function. In the example above $2 \times 2 = 4$. So, this process can only produce a cubic curve in trivial cases where either the original curve or the reparameterization function is linear.