A bit of confusion with central idea of "running" coupling constants An effective quantum field theory of a single scalar field $\phi$ is described by an action, $S(\phi,\{g_n\})$ where $\{g_n\}$ denote the coupling constants of the theory. The corresponding path-integral $$Z=\int\mathcal{D}\phi\exp[iS]$$ is usually defined with a finite UV cut-off $\Lambda$. When the cut-off is lowered from the scale $\Lambda$ to $\Lambda^\prime=\frac{\Lambda}{b}$ ($b> 1$) by integrating out the fast Fourier components with momenta between $\Lambda^\prime$ and $\Lambda$, we obtain another lower energy effective action, $$S^\prime(\phi,\{g^\prime_n(\Lambda, \Lambda^\prime)\}).$$ So far, I have tried not to be notationally inconsistent or incomplete. Assuming the description above is all fine, I can understand that the coupling constants will "run" with the lowering of the cut-off. The coupling constants $\{g_n\}$s of the old effective theory (with cut-off $\Lambda$) will change into $\{g_n^\prime\}(\Lambda,\Lambda^\prime)$s for the new effective theory (with the lower cut-off $\Lambda^\prime$). Please correct me for mistakes.
Now consider the initial effective theory $S(\phi,\{g_n\})$ with the fixed cut-off $\Lambda$. By fixed I mean, unlike the description given above, now I am neither interested in lowering the cut-off $\Lambda$ nor interested in finding a lower energy effective theory. Instead, if I am interested in finding how the coupling constants $\{g_n\}$ of my old theory (with the fixed cut-off $\Lambda$) vary with energy $E$ in the range $0<E<\Lambda$. How should I proceed and what do I expect?
 A: If you fix the cutoff, in a certain sense, the coupling constants don't run. But it's better to talk about concrete things to it's clear what's going on.
Fix parameters in the Lagrangian
If you define $g_i$ to be the parameters that you have in the Lagrangian at a certain cutoff $\Lambda$, then no, they do not depend on the energy of your experiment. Those are fixed once and for all.
Define a coupling as a physical correlator
A coupling constant may be defined as the value of an $n$-point correlator at a certain energy $E_0$. More precisely, say you have a three-point coupling $g\,\phi^3(x)$, you can define $g$ as
$$
g = \langle \phi(p_1)\phi(p_2)\phi(p_3)\rangle\big|_{|p_1|=2|p_2|=2|p_3|=E_0}\,,
\tag{1}$$
or something like that. So again, if you fix $E_0$ to a value that you like, $g$ doesn't run. What's true is that the tree-point correlator written above depends nontrivially on the energy $E_0$. Which is not surprising at all. But if $E_0$ is fixed, of course it's constant.
In even more concrete terms: if you compute an amplitude $\mathcal{A}(g,E)$ at energy $E$, and then you compute the same amplitude $\mathcal{A}(g,E')$ at higher energy $E'$, the $g$ that goes in is the same number.
So why does the coupling run?
Saying that the coupling remains unchanged is kind of moot, precisely because you can choose at what scale to define it. That means that the value of $g$ itself tells you literally nothing about the physics.
What matters for the physics at energy $E$ are the values of the correlators around the same energy scale. So if someone asks you "is this theory strongly coupled?" or "is this theory close to a free theory?" you might want to look at the correlators at those energy scales.
So, for any computations, even at $10^{10^{10}}\,\mathrm{GeV}$, you can keep you coupling defined at the scale $10\,\mathrm{eV}$ and get the correct value. But the physics of the quantum system is more transparent if you redefine $g$ to be $g$ in $(1)$ at the scale of $E_0=10^{10^{10}}\,\mathrm{GeV}$. For instace, some theories are asymptotically free. But, if you fix $g$ at low energies, $g$ is order one and the amplitudes that you compute with it are close to a free theory when $E$ is big.${}^1$ That's kind of perverse.
So it's not the coupling itself that runs, is that we force ourselves to redefine it into a sensible energy range so that there is some meaning attached to it.

$\;{}^1$Nevermind that in real life you won't be able to make a computation with a large coupling because we can deal with QFT only by using perturbation theory, for most situations. For the purpose of this answer, assume that we are all-powerful and can compute all sorts of stuff.
