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
$\,M_{m,n}$ .