Different frequencies and the same $n-$revolutions The frequency $\nu$ it is defined as:
$$\nu=\frac1T \tag 1$$
where with $1$ indicate one cycle and $T$ the period, i.e. when a material point completes a complete circle. If I want generalize the $(1)$, I write:
$$\nu^*=\frac{n^*}{\Delta t}$$
where $n\in \Bbb N\smallsetminus\{0\}$ is the number of revolutions and $\Delta t$ it is the time interval required to complete $n-$revolutions.
If the
$$\bbox[5px,border:2px solid #138D75]{\nu=\nu^*=\text{constant} \implies\frac1T = \frac{n}{\Delta t} \iff n=\frac{\Delta t}{T}}\tag 2$$

If I have $\nu\ne \nu^*$ do it exists a relation (or an practical example) where I can get a different relation of the $(2)$ to obtain the same $n-$revolutions?

Addendum (further explanation):
If we have $\nu\ne \nu^*$ obviously we will have
$$\frac{n}{\Delta t}\ne \frac{n^*}{\Delta t^*} \implies $$
$$\Delta t\neq \Delta t^* \,\text{ or } \, n\neq n^*\tag 3$$
If it is  $n\neq n^*$ the case is closed.  If $$\bbox[5px,border:2px solid #BA4A00 ]{\Delta t\neq \Delta t^*  \stackrel{?}{\implies} n=Kn^*, \, K\in\Bbb R} \tag 4
$$

When does this occur the case $(4)$ i.e. $\nu\neq \nu^* \implies n\propto n^*$, and in particular for $K=1$ i.e. $n= n^*$?

 A: I don't see how $\nu$ and $\nu^*$ can be different from each other. You are defining  $\nu*$ as $n/\Delta t$, but clearly the time for one cycle, the periodic time, $T$, is $\Delta t/n$. Therefore $\nu^* =1/T =\nu$.
Have I misunderstood what you are asking about?
A: It is not 100% clear what exactly is the question... however there is one apparent issue which is probably the core of the problem: one discusses integer number of revolutions/cycles, while allowing the time to be continuous.
Thus, the following math makes sense only when the frequencies are commensurate, i.e., their ratio is the ratio of integers:
$$
\frac{\nu}{\nu^*}=\frac{n}{n^*}, \quad n, n^*\in \mathbb{N} \backslash \{0\}
$$
Otherwise, for the given time interval, one of the rotating objects may not have completed an integer number of rotations/cycles.
This problem could be resolved by treating the number of rotations as a continuous variable, $n \in \mathbb{R}$ - in more common physics langauge it means defining the phase of rotations/cycles,
$$\phi=2\pi n, \quad \phi \in \mathbb{R},
$$ and defining the frequency via this phase:
$$
\omega=2\pi\nu = \frac{\Delta\phi}{\Delta t}
$$
A: I will take a bit of a guess of what you wanted to ask, if I'm incorrect, I'll remove this answer again, just let me know.
So, you have two signals with $\nu_1 \neq \nu_2$, so different periods $T_1$ and $T_2$ (allow me to introduce the subscript notation $\nu_{1}$, $\nu_{2}$, it feels a bit more natural than using the a (un)starred $\nu$ and $\nu^*$). But you could equally well find their respective frequencies by counting the number of oscillations and dividing this by the time interval you used for counting. Since the frequencies differ, you have the relationship
$$ \nu_1 = \frac{n_1}{\Delta t_1} \neq \frac{n_2}{\Delta t_2} = \nu_2.$$
You want to know how to make sense of different $n_{i}$ and $\Delta t_{i}$ with $i = 1,2$. The key answer: as an experimentalist, you choose a fixed $\Delta t$ and then look for the resulting $n$. Since the motion doesn't change over time, you will find exactly the period $T$ by this procedure. You can also decide to choose a fixed $n$ and then determine $\Delta t$.
I think it's best to go to an example, it is really quite a simple matter. Say you want to count your heart rate. You put your hand on your wrist and get a stop watch ready. You set an alarm to one minute (a time that you, conducting this experiment, just chose yourself) and then you start counting how often you feel your pulse. 1,2,3,... When you reach 45, the alarm goes off, one minute has passed. So you counted $n=45$ heart beats in $\Delta t = 60s$. With that, you can calculate the heart rate to be
$$\nu_{\textrm{heart}} = \frac{n}{\Delta t} = \frac{45}{60s} = \frac{3}{4} \textrm{Hz},$$
leading to a heart period of $T_{\textrm{heart}} = 4/3 s$. You could have equally well calculated for 2 minutes or 10 or half a minute (although the shorter your time interval, the less accurate your measurement becomes). The important bit is the ratio $n/\Delta t$.
So if you were to calculate your heart rate and your next door neighbor's, you can both use the same alarm clock and one of you counts $n_1 =45$ and the other $n_2=40$ and the two of you have the heart rates
$$\nu_1 = \frac{3}{4}\textrm{Hz} = \frac{\nu_1}{\Delta t} \neq \frac{\nu_2}{\Delta t} = \frac{2}{3} \textrm{Hz} =\nu_2.$$
But maybe you used different procedures, say your neighbor measured for $1.5$ min and came to a total of 60 heart beats. You can still compare, and $\nu_2$ is now
$$\nu_2 = \frac{60}{90\textrm{s}} = \frac{2}{3}\textrm{Hz},$$
so nothing changed. The important bit is that there must not be any integer relation between $\Delta t_1$ and $\Delta t_2$, so there is no integer relation between $n_1$ and $n_2$, either. You may have some sort of rounding error, but the effect of that gets reduced when you choose $\Delta t_1$ and $\Delta t_2$ sufficiently large.
You asked about math. A simple math reason is that there is a linear relation between $n$ and $\Delta t$, namely
$$\nu = \frac{1}{T} = \frac{n}{\Delta t} \quad \iff \quad n = \frac{\Delta t}{T},$$
and this is what the previous answer already contained.
I don't know if there is a much deeper math answer that's useful. I guess the deepest answer I can give is that the rational numbers are an equivalence class $\mathbb{Q} = (\mathbb{N}_0 \times \mathbb{N})/_{\sim}$, where two integer pairs $(a,b)$ are equivalent to $(c,d)$ if there are integers $m,n \in \mathbb{N}$ such that $ma = nc$ and $mb = nd$, because then
$$\frac{a}{b} = \frac{ma}{mb} = \frac{nc}{nd} = \frac{c}{d}$$
in typical $\mathbb{Q}$ arithmetic. Don't know if that helps much, but it's basically the mathematical reason why we can reduce fractions, and there isn't anything deeper happening (although admittedly $\Delta t$ is a real number).
A: I do not understand what the confusion is. The way I see it, things are simple.
Over a period of time $\Delta t$ under constant rotational speed $\omega$ the object accrues the following number of revolutions
$$ n = \frac{\omega \Delta t}{2 \pi} $$
As a shorthand we use frequency $\nu = \frac{\omega}{2\pi}$ above for $n = \nu \Delta t$, but that is just so we don't have to write the $2\pi$ every time. Note the units of $\omega$ are $\text{radians/second}$.
So you want the same number of revolutions you have to change both the time frame and the rotational speed
$$ n = \frac{\omega_1 \Delta t_1}{2 \pi} = \frac{\omega_2 \Delta t_2}{2 \pi} $$
or
$$ n = \nu_1 \Delta t_1 = \nu_2 \Delta t_2 $$
So if you agree up to this point, what is the point of confusion here?
As far as period $T$ it is defined as the time for one revolution, so take $1 = \frac{\omega T}{2 \pi}$ and solve for $T$
$$ T = \frac{ 2 \pi}{\omega} = \frac{1}{\nu} $$
A: I use those equations
$$\omega=2\,\pi\,\nu~,\nu=\left[\text{Hz}\right]$$
$$\omega=\frac{n\,\pi}{30}~,n=\left[\frac{1}{\text{min}}\right]~$$
$$T=\frac 1\nu~,T=[s]$$
$\Rightarrow$
$$2\,\pi\,\nu=2\,\pi\frac{1}{T}=\frac{n\,\pi}{30}~,T=[\text{min}]$$
thus:
if the frequencies are equal $~\nu_1=\nu_2~\Rightarrow$ $~,n_1~$ must be equal $~n_2~$ and $~,T_1$ must be equal $~T_2$
Edit
$$\nu_1=\frac{1}{T_1}=\frac{n_1}{60}$$
$$\nu_2=\frac{1}{T_2}=\frac{n_2}{60}$$
case I
$~\nu_1=\nu_2$
$\Rightarrow$
$$\frac{1}{T_1}=\frac{1}{T_2}~,T_1=T_2$$
and
$$\frac{n_1}{60}=\frac{n_2}{60}~,n_1=n_2$$
Case II
$~\nu_1\ne\nu_2$
$\Rightarrow$
$$\frac{n_1}{n_2}=\frac{\nu_1}{\nu_2}=K~,\text{where}~ K> 0 \in\mathcal{R}$$
this is a constraint equation thus you can choose either $~n_1~$ or $~n_2$ not both
A: Your definitions of $\nu^{*}$ is no generalization of the frequency $\nu$. These are equivalent formulations. Because time is continuous you can always reduce the elapsed time to one cycle.
A: I don't know what this question is supposed to be useful for. But the answer is pretty simple. We have the functional relationship
$$n:\Bbb R\times\Bbb R\to \Bbb R$$
$$n:(\nu,\Delta t)\to \nu\cdot\Delta t$$
Since you require
$$n=Kn^*$$
and you know that $\nu$ and $\nu^*$ have to satisfy $\nu\ne\nu^*$
we substitute our function $n(\nu,\Delta t)$ and rearrange the above relation so that the known quantities are on the right hand side
$$\frac{\Delta t}{\Delta t^*}=K\frac{\nu^*}{\nu}$$

If I have $\nu\ne \nu^*$ do it exists a relation (or an practical example) where I can get a different relation of the $(2)$ to obtain the same $n-$revolutions?

If the number of revolutions need to be the same, we have $K=1$ and so,
$$\frac{\Delta t}{\Delta t^*}=\frac{\nu^*}{\nu}$$
Put your specific numbers $\nu\ne\nu^*$ in there, and you get the relation that $\Delta t$ and $\Delta t^*$ have to satisfy.

$$\Delta t\neq \Delta t^*  \stackrel{?}{\implies} n=Kn^*, \, K\in\Bbb R$$ When does this occur the case $(4)$ i.e. $\nu\neq \nu^* \implies n\propto n^*$, and in particular for $K=1$ i.e. $n= n^*$?

The case $K=1$ has already been answered above. If $K\in\Bbb R$ is arbitrary and fixed, the relation for $\Delta t$ and $\Delta t^*$ you have asked for is
$$\frac{\Delta t}{\Delta t^*}=K\frac{\nu^*}{\nu}$$
If $K$, or $\nu$, or $\nu^*$ is completely arbitrary, there are no constraints on $\Delta t$ and $\Delta t^*$, other than being nonzero (and presumably non-negative). Just choose anything you like, for example point your finger blindly to some entry in the phone book, take the square root of it, multiply it by $\pi$, and so forth.
Caveat: due to the fact that $n$ and $n^*$ are nonzero integers, $K$ can only be, mathematically, a rational number, not an arbitrary real, i.e.
$$K\in\Bbb Q$$
However, since any real number can be approximated by a rational arbitrarily accurately, and on the other hand measurements of $\nu$ and $\nu^*$ can't cover the reals anyway, $K$ can also be assumed an arbitrary real for all practical (physical) purposes.
