When do equations represent the same curve?

Consider the simpler problem of two parametrized curves $(x_i(t),y_i(t))$ that start at $(0,0)$ at time $t=0$, and equality of trajectories up to reparametrization, which is stronger than equality of point sets and a more natural condition as it is local (up to matching of starting points). Heuristically, and to some extent rigorously, there is a usable criterion.

We want that when $x_1 = x_2$, then $y_1 = y_2$ so that the bivariate function $x_1(t)-x_2(s)$ divides $y_1(t)-y_2(s)$ (and vice versa) in a suitable ring of functions. Their ratio is an invertible function with positive values, at least for nonzero $s,t$ near $0$. In fact we need it to be positive only for nonzero $(s,t)$ at which $x_1(s)=x_2(t)$ or the same for $y$.

Example: parabola and half-parabola.

Curve A is $(t,t^2)$.

Curve B is $(s^2,s^4)$.

$x_1(t)-x_2(s) = t - s^2$

$y_1(t)-y_2(s) = t^2 - s^4$

Ratio is $(t + s^2)$

This is positive near the locus where $x_1(t)=x_2(s)$ (namely $t=s^2$). On the locus where $y_1(t)=y_2(s)$ (namely $t^2=s^4$), this is positive for $t>0$ and negative for $t < 0$. The positivity condition knows which half of the parabola is curve B! That is a good sign that this is either the complete answer to the simplified problem, or on the right track.

Finding an intersection point of two parametric curves OR detecting a difference between the curves is simpler than the general problem of curve intersection. Take a point on one curve, solve for the parameter values that would place its $x$ coordinate on the other curve, and test whether the $y$ coordinates are the same. For algebraic parameterizations this calculation can be done exactly.

For the point-set equality problem, locate the zeros of the $st$ ratio. These parameter values segment the two curves into arcs. Then there is a combinatorial problem of orienting and matching (by the procedure given above) identical arcs of the two curves, and testing whether both curves are covered by the matched arcs.

Disclaimer. This answer is more like a question. Since the (+100) bounty has been raised by the same author, it will not be attached to this answer. But if someone can prove my parametrization conjecture (see below), that would be great. (Update. The bounty hasn't been awarded to anyone)

Because we are going to calculate definite integrals, it is important to have closed intervals for the parameters $t$ . For the example as presented in the question: $$ \mathbf c_1(0) = (1,0) \quad ; \quad \mathbf c_1(1) = (0,1) \\ \mathbf c_2(0) = (1,0) \quad ; \quad \mathbf c_2(1) = (0,1) $$ So we are lucky: if both parameters $t$ are allowed to be restricted to the interval $[0,1]$ then the two curves at least have the same end-points.
Consider the expression $\left[x(t)y'(t)-y(t)x'(t)\right] dt$ . It is twice the area of an infinitesimal triangle with vertices at $(0,0)$ , $(x(t),y(t))\,$ and $\,(x(t+dt),y(t+dt))$ : $$ 2 \times \mbox{area}\,\Delta = det\begin{bmatrix} x(t) & x(t+dt) \\ y(t) & y(t+dt) \end{bmatrix} = \left[x(t)\frac{y(t+dt)-y(t)}{dt} - y(t)\frac{x(t+dt)-x(t)}{dt}\right] dt $$ Now we are going to calculate the following quantities, with the end-points $\,(a,b) = (0,1)\,$ and $\,m,n\,$ positive or zero integers. Integration over an area instead of an arc length is to be preferred for some good reasons (: e.g. suppose that the curve is traversed back and forth at some places). $$ M_{m,n} = \int_a^b x(t)^m y(t)^n \left[x(t)y'(t)-y(t)x'(t)\right] dt $$ The quantities $M_{m,n}$ , not at all by coincidence, are similar to variances (in statistics terms) or moments of inertia (in physics terms). It is conjectured that the outcome of these integrals is independent of any parametrization. This may be called a Parametrization Conjecture.
It is assumed to be true in the sequel. Can someone prove or disprove?

MAPLE has been invoked to save time and effort. Definitions for the example as given in the OP's question and first few steps:

x1(t) := (1 - (2 - sqrt(2))*t - (sqrt(2) - 1)*t^2)/
         (1 - (2 - sqrt(2))*t + (2 - sqrt(2))*t^2);
y1(t) := (sqrt(2)*t - (sqrt(2) - 1)*t^2)/
         (1 - (2 - sqrt(2))*t + (2 - sqrt(2))*t^2);
x2(t) := (1-t^2)/(1+t^2); y2(t) := 2*t/(1+t^2);
x1d(t) := simplify(diff(x1(t),t)); y1d(t) := simplify(diff(y1(t),t));
x2d(t) := simplify(diff(x2(t),t)); y2d(t) := simplify(diff(y2(t),t));
M00 := int(x1(t)*y1d(t)-x1d(t)*y1(t),t=0..1);
N00 := int(x2(t)*y2d(t)-x2d(t)*y2(t),t=0..1);
verify(M00,N00,equal);
                           true

For the example at hand, the lowest order (area) moments are indeed exactly the same for the two parametrizations: $$ M_{0,0} = \frac{\pi}{2} \\ M_{1,0} = 1 \quad ; \quad M_{0,1} = 1 \\ M_{2,0} = \frac{\pi}{4} \quad ; \quad M_{1,1} = \frac{1}{2} \quad ; \quad M_{0,2} = \frac{\pi}{4} \\ M_{3,0} = \frac{2}{3} \quad ; \quad M_{2,1} = \frac{1}{3} \quad ; \quad M_{1,2} = \frac{1}{3} \quad ; \quad M_{0,3} = \frac{2}{3} \\ M_{4,0} = \frac{3\pi}{16} \quad ; \quad M_{3,1} = \frac{1}{4} \quad ; \quad M_{2,2} = \frac{\pi}{16} \quad ; \quad M_{1,3} = \frac{1}{4} \quad ; \quad M_{0,4} = \frac{3\pi}{16} \\ M_{5,0} = \frac{8}{15} \; ; \; M_{4,1} = \frac{1}{5} \; ; \; M_{3,2} = \frac{2}{15} \; ; \; M_{2,3} = \frac{2}{15} \; ; \; M_{1,4} = \frac{1}{5} \; ; \; M_{0,5} = \frac{8}{15} $$ And so on and so forth. In this way it may be confirmed, step by step, that the two parametrizations represent one and the same curve. It remains unsatisfactory that we cannot establish everything in one step, though: MAPLE could not calculate for the general expression $\,M_{m,n}$ .

It would be easier for you if you have a nice set of definitions to understand your problem. So here I drop some of them hoping they will let you arrive at the correct answer. If you still have problems just ask in the comments.

By a $C^1$ path in $\Re^n$ is meant a continuously differentiable function $\gamma:[a,b] \rightarrow \Re^n$.

The $C^1$ path $\gamma:[a,b] \rightarrow \Re^n$ is said to be smooth if $\gamma' (t) \not= 0$ for all $t \in [a,b]$.

Having defined that, suppose that $\alpha:[a,b] \rightarrow \Re^n$ and $\beta:[c,d] \rightarrow \Re^n$ are two $C^1$ paths that are "geometrically equivalent" in the sense that they have the same initial point and the same terminal point, i.e., $\alpha(a)= \beta(c)$ and $\alpha(b)= \beta(d)$ respectively. Then we say that the path $\alpha:[a,b] \rightarrow \Re^n$ is equivalent to the path $\beta:[c,d] \rightarrow \Re^n$ if and only if there exists a $C^1$ function $$\psi:[a,b] \rightarrow [c,d]$$ such that $\psi([a,b])=[c,d], \alpha = \beta \circ \psi$ and $\psi'(t)> 0$ for all $t\in[a,b].$

The set $C \subseteq \Re^n$ is called a curve if and only if it is the image of a smooth path $\gamma$ which is one-to-one. And any one-to-one smooth path which is equivalent to $\gamma$ is then called a parametrization of $C$.

Finally if $\vec{x}=\gamma(t) \in C$ then $\vec{T}(\vec{x}) =\frac{\gamma'(t)}{|\gamma'(t)|}$ is a unit tangent vector to $C$ at $\vec{x}$, and $\vec{T}(\vec{x})$ is independent of the chosen parametrization $\gamma$ of $C$. Such a continuous mapping $T:C \rightarrow \Re^n$ is called an orientation for $C$. So an oriented curve is then a pair $(C,T)$ or just $C$. And $-C$ is the same geometric curve with the opposite orientation $(C,-T)$.