$\def\RR{\mathbb{R}}
\def\vr{\mathbf{r}}
\def\vv{\mathbf{v}}
\def\vnull{\mathbf{0}}
\def\ss{\subseteq}
\def\To{\rightarrow}
\def\p{\pi}$Consider a curve in $\RR^n$ given by
$\vr(t) = (x_1(t),\ldots,x_n(t))$ for $t\in I\ss\RR$.
Suppose $\vr$ is differentiable and non-self intersecting.
(The restriction of non-self intersecting can be relaxed, but introduces technicalities that we wish to avoid.)
Let $\vv(t) = \vr'(t)$ and further suppose that $\vv(t)\ne \vnull$ for any $t$.
The arclength along this curve will be given by
\begin{equation}s=\int_{t_0}^t |\vv(t)|dt,\tag{1}\end{equation}
where $t_0\in I$ is some reference value for the parameter $t$ corresponding to an arclength of $s=0$.
(Note that $t<t_0$ corresponds to a negative arclength.
This sign agrees with the orientation of the curve.)
By assumption, $|\vv|>0$.
Thus, $s$ is a strictly monotone increasing function of $t$ and so will have an inverse function, $t=t(s)$.
Thus,
$\vr(s)\equiv\vr(t(s))$,
where $s\in I'\ss\RR$,
will be this same curve parametrized by arclength.
The interval $I'$ can be determined from (1).
After this procedure has been done, it should be clear that two parametrizations of the same curve,
$\vr_1(s)$ and $\vr_2(s)$,
can differ in their representation only through the following transformation:
$s\To ks+s_0$,
where $k=\pm 1$.
That is, the curves can only differ by a shift in $s$ or by having opposite orientations.
(It is also possible that after this transformation further transformations on $I'$ may be allowed and necessary, depending on the periodicity of the components of $\vr(s)$.)
A quick way that one can distinguish different curves is by examining the intervals $I'$.
For example, if the intervals are finite subsets of $\RR$ and are different in length, then the curves are not the same. (They do not have the same length.)
Examples:
Let
\begin{align*}
\vr_1(t) &= (\sin t,\cos t),
& t\in[0,2\p) \\
\vr_2(t) &= (\cos 2t,\sin 2t),
& t\in[0,\p) \\
\vr_3(t) &= \left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right),
& t\in\RR
\end{align*}
(The last example is @Doug M's from the comments.)
It is a straightforward exercise to show that
\begin{align*}
\vr_1(s) &= (\sin s,\cos s),
& s\in[0,2\p) \\
\vr_2(s) &= (\cos s,\sin s),
& s\in[0,2\p) \\
\vr_3(s) &= \left(\cos s,\sin s\right),
& s\in(-\p,\p)
\end{align*}
Here we have let $t_0=0$ for each parametrized curve for simplicity.
Note that
$\vr_1(-s+\p/2) = \vr_2(s)$.
Under this transformation, $I_1=[0,2\p)\To[-3\p/2,\pi/2)$.
Using periodicity, $I_1\To[0,2\p)$.
Thus, the first and second parametrized curves correspond to the same graph.
Using periodicity we find
$I_3=(-\p,\p)\To[0,2\p)\setminus\{\p\}$.
Thus, the third parametrized curve corresponds to a different graph.
As an exercise, using this procedure one should be able to determine that the following curves correspond to the same graph.
\begin{align*}
&(\sin t,\cos t),
& t\in[0,\p] \\
&(\cos 2t,\sin 2t),
& t\in[-\p/4,\p/4] \\
&\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right),
& t\in[-1,1]
\end{align*}
Addendum added to react to the comment question of the original poster
The computations can be simplified by temporarily assuming that one endpoint of the hypotenuse has Cartesian coordinates $(0,0)$ and the other endpoint has Cartesian coordinates
$\displaystyle (R,0) ~: R = \sqrt{A^2 + B^2} = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}.$
Then, at this stage, all that is missing is the Cartesian Coordinate of the $3$rd vertex. The easy way is to construe the line segment going from the origin to the 3rd vertex as a vector of length $R\cos(\alpha)$ and direction $(\alpha)$
Here, I am making the standard assumption that the positive directions of angles are in a counter-clockwise direction.
If you then take scratch paper, and draw the corresponding diagram, you will see that the Cartesian coordinates of the $3$rd vertex are therefore
$$\left[R\cos^2(\alpha), ~~R\cos(\alpha)\sin(\alpha)\right]. \tag1 $$
All that remains it to rotate and then shift the diagram, and compute the resulting affect on the Cartesian coordinates of the $3$rd vertex, which is currently shown in (1) above.
Under the initial temporary assumption, the slope of the line $\left(\overline{AB}\right)$ is $0$.
The actual slope needs to be
$$M = \frac{y_1 - y_0}{x1 - x_0}.$$
So,
If $M > 0$, then let $\theta = \arctan(M).$
If $M < 0$, then let $\theta = (\pi) - \arctan(-M) \implies $
$\pi/2 < \theta < \pi.$
An easy way of then performing the rotation is to repeat the procedure that was initially used to compute the Cartesian coordinates shown in (1) above. Instead of construing the pertinent leg of the triangle to be a vector of magnitude $R\cos(\alpha)$ and direction $(\alpha)$, simply construe the pertinent leg of the triangle to be a vector of magnitude $R\cos(\alpha)$ and direction $(\alpha + \theta)$.
Then, the rotational affect is to alter the Cartesian coordinates, shown in (1) above. Instead, they become
$$(P,Q) = \left[R\cos(\alpha)\cos(\alpha + \theta), ~~R\cos(\alpha)\sin(\alpha + \theta)\right]. \tag2 $$
So, in (2) above, $(P,Q)$ represents the Cartesian coordinate of the $3$rd vertex, after the rotational adjustment, but before the shift adjustment.
The shift adjustment is straightforward. Assuming that $(0,0)$ is shifted to $(x_0, y_0)$, you have that
$$(P,Q) ~~\text{is shifted to}~~ (P + x_0, Q + y_0).$$
Addendum
The OP (i.e. original poster) indicated a flaw in my response. What happens if $(x_1 - x_0)$ equals $0$. In that case, the attempt to compute of $M$ will fail, so the strategy used to determine $(\theta)$ will fail.
That situation represents a vertical hypotenuse, which represents a rotation of $(\pi/2)$. So, in that case, set $(\theta)$ to $(\pi/2)$.
Best Answer
First, notice that by reparameterizing the Grandi roses by replacing $t$ with $2 t$ (so, doubling the speed), we can write the curves respectively as graphs of polar functions $\rho, \textrm{P}$ in the angular variable (which I'll still call $t$):
\begin{align} \rho &:= 1 + r \sin k t \\ \textrm{P} &:= R \sin k t . \end{align}
(1) This viewpoint quickly explains both of the differences we notice in the plots with $k$ even: First, we see that if we plot both $\rho, \textrm{P}$ over a full period (an angular interval of $2 \pi$), the asymmetry of the plots with $k$ even disappears: With the original parameterization, we only trace for half a period of $\textrm{P}$, which is anyway less natural. Plotting both curves over a full period for even $k$ gives this more symmetric graph (for $r = 1 / 3, k = 6$).
We can also see immediately which this issue didn't occur for odd $k$, even with the slower parameterization. Expanding using the usual sum formula gives $$\textrm{P}(t + \pi) = R \sin k(t + \pi) = (-1)^k R \sin k t = (-1)^k \textrm{P}(t).$$ But the polar coordinates $(t + \pi, \alpha)$ and $(t, -\alpha)$ represent the same point, which (for $r > 0$) tells us exactly that the parameterization $\textrm{P}(t)$ traces the graph of $\textrm{P}$ with period $\pi$ iff $k$ is odd. So, for $k$ odd we still get the complete graph by plotting it over an angular interval of length $\pi$, or equivalently over an angular interval of length $2 \pi$ (rather than $4 \pi$) in the original parameterization of the Grandi rose. The same identity also immediately more-or-less explains the more essential difference between the situations for $k$ even and odd, namely that for $k$ even (but not odd) the lobes of the curve $\textrm{P}$ extend outside the graph of $\rho$.
(2) We can see that for $r = 1$, $\rho$ has a minimum of zero---and so its graph intersects the origin---at integer multiples of $\frac{\pi}{k}$. When $r < 1$, $\rho(t) \geq 1 - r > 0$, in which case this behavior doesn't occur.
(3) It's really not clear to me what is meant here. But NB for $r < 1, k > 1$, the polar graphs of $\rho, \textrm{P}$ (as subsets of $\Bbb R^2$) are topologically inequivalent: The polar graph of $\rho$ is topologically equivalent to a circle, whereas the polar graph of $\textrm{P}$ is topologically equivalent to a bouquet of $k$ circles (for $k$ odd) and of $2 k$ circles for $k$ even. For $r = 1$, the polar graphs are topologically equivalent iff $k$ odd.
(4) Notice that we can preserve some of the qualitative behavior of the examples if we replace $\sin kt$ with any other odd function with period $2 \pi$. In particular, taking the first two interesting terms of the Fourier series of any such function gives pairs
\begin{align} \tilde \rho &:= 1 + r (\sin k t + a \sin 3 k t ) \\ \tilde{\textrm{P}} &:= R (\sin k t + a \sin 3 k t) . \end{align}