Is it possible to make the electrons position stationary at a fixed position? Is it possible for an electron to just be in a fixed position?
I can't really think of a good experimental set-up to explain what I'm trying to say but for example:
Let's say we have an electron accelerator that shoots electrons with different kinetic energies. Then if we set up an $E$-field such that it will do work on ejected electrons until it loses all the kinetic energy and the $E$-field turns off at a perfect time such that electron's kinetic energy is zero. Ignoring the technical difficulty, wouldn't it be possible to give out all the kinetic energy of an electron so that it becomes stationary? But then, I think Heisenberg's uncertainty principle says we can't. Or maybe I understanding the uncertainty principle wrong? Can position of this electron (or any object at quantum scale) be fixed in position if there is no interaction (say in a vacuum) what so ever?
 A: 
Heisenberg's uncertainty principle says we can't

And we can't - even if there are no interactions whatsoever, the electron will not remain still.
See also zero point energy. It is not possible for a quantum system to have $0$ kinetic energy, because of Heisenberg's uncertainty principle.
A: 
Is it possible to make the electrons [sic] position stationary?

Yes, depending on what you mean by "stationary."
In quantum mechanics, "stationary states" are eigenstates of the Hamiltonian.
If the electron is in a stationary state, then the expectation value of its position does not depend on time.
For example, when $|\Psi_E\rangle$ is a stationary state such that $i\partial |\Psi_E\rangle/\partial t = H|\Psi_E\rangle = E|\Psi_E\rangle$, then we have:
$$
x_E(t) \equiv \langle \Psi(t)| \hat X |\Psi(t)\rangle
$$
and
$$
\frac{dx_E}{dt} = \frac{\partial \langle \Psi(t)|}{\partial t}\hat X|\Psi(t)\rangle
+
\langle \Psi(t)|\hat X\frac{\partial |\Psi(t)\rangle}{\partial t}
$$
$$
=iE\langle \Psi(t)| \hat X |\Psi(t)\rangle - iE\langle \Psi(t)| \hat X |\Psi(t)\rangle
$$
$$
=0
$$

As a secondary proof, consider a common case, where the stationary state can be written as:
$$
\Psi(x,t) = e^{-iEt}\phi(x)\;,
$$
where $\phi(x)$ is real and goes to zero at $\pm\infty$.
In this case we can easily see that the expectation value of the momentum is zero:
$$
\langle \hat P \rangle = -i\int dx \phi(x)\frac{d\phi}{dx}
=\frac{-i}{2}\int dx \frac{d|\phi|^2}{dx} = |\phi|^2(\infty) - |\phi|^2(-\infty) = 0 
$$
For any Hamiltonian of the form
$$
H = \frac{\hat P^2}{2m} + V(x)\;,
$$
one can show that
$$
\frac{1}{m}\langle\hat P\rangle = \frac{d}{dt}\langle\hat X\rangle\;.
$$
I.e., we again see that the time derivative of the expectation value of the position is zero in a stationary state.

UPDATE
OP has asked in the comments about an electron in a position eigenstate, with fixed position $x_1$. Such a state in the position basis looks like:
$$
\chi(x) = \delta(x - x_1) = \int\frac{dp}{2\pi}e^{ip(x-x_0)}\;,
$$
which is not a physical state, and clearly has contributions from all momenta (see, e.g., the far RHS of the above equation).
If one tries to create a physical/normalizable state, like a gaussian wave packet, one will find that the electron can not be normalized in both position and momentum any better than expected by the known rule:
$$
\sigma_x \sigma_p \ge \hbar/2
$$
A: You start with a localized electron that's moving with some nonzero expected value of momentum, and thus nonzero kinetic energy. Then you remove all this momentum using an external E-field.
The catch is in the localization. If the electron is localized, its momentum is not definite (Heisenberg's uncertainty principle). Instead there's a probability distribution of different momenta, with some mean value that you remove by the external E-field. This just shifts the distribution to be around $\vec p=0,$ while the uncertainty is still preserved.
Now, to get the expected value of kinetic energy you would square all these momenta and integrate over them with a particular weight function (proportional to the probability density). Since there are nonzero momenta in the distribution, your expected value of kinetic energy will be nonzero. Meanwhile, the electron's distribution of positions will spread out as time passes (assuming free space).
A: You are trying to apply classical intuition to a quantum mechanical system.  You have correctly understood the way the Heisenberg Uncertainty principle is usually described "You can't simultaneously fix a particle's momentum and position", but you're missing the reasons for that principle and that's leading you into a loop of seemingly-reasonable "but what if..."
This is common!  Welcome to Quantum Mechanics! It's weird and counter-intuitive until you've done a lot of practice with it, seen the proofs, and run down a few scenarios.  We're here to help.
So let's look at Heisenberg a bit more closely. It's actually a mathematical statement and applies to more pairs than just position & momentum (or speed). (The detailed math requires comfort with integral calculus.) Assuming you don't want to go through the mathematical details but do want to improve your intuition about why this won't work, there are just a few pieces you need to take on faith (or research until you're satisfied).
Use these when you're trying to build a mental model of what's likely to happen:

*

*Every "particle" in quantum mechanics is represented by a function which is distributed in space. This function is basically a probability distribution.

*The "most likely position" of the particle is the place where the function is highest. Other "likely positions" are places where the function is significantly larger than zero, but not as high as the "most likely" position. You get quantum tunneling / teleportation when you have a function with large values in two different places and a gap of zero or near-zero in between: The particle can be in either location, but can't exist between those spots.

*And this is the crazy one: The "momentum" if the particle is the slope of the probability distribution.  So if the distribution is locally steep / rapidly changing, that particle's moving.  If it's pretty flat, the particle's stationary.

(Strict mathematical definitions & proofs don't completely line up with those statements, but they're still a good approximation until you're comfortable with the math. You can absolutely break the above model if you push it too hard, but as long as you're not focused on edge cases, it will serve you better than classical intuition built up by playing catch with baseballs.  For example, in hand-waving arguments, we can call the probability distribution the "wavefunction", but the wavefunction and probability distribution are formally, mathematically distinct.)
So let's see what this hand-wavy model has to say about your decelerate-an-electron-to-zero idea:

*

*You want to fix the position of the electron at some location.  Let's call that spot the origin, 0, for simplicity.  You want the electron to be definitely there, so the electron's distribution function (or wavefunction) is going to be sharply peaked at 0.  If that's what your wavefunction looks like, everyone will agree the electron is at the origin.

*However, because we want the electron's distribution to be sharply peaked, there are large slopes involved: That distribution rises from zero to 1 and falls back down to zero instantly.  That means the momentum is not known.  It appears to be infinite... or maybe negative infinite... or maybe zero?  Depending on which side of the spike you measure, or whether you average over the whole spike, or if the peak isn't quite symmetric or... So the uncertainty in your momentum is huge.

*What about starting from the other end? You've stopped the electron and so you're guaranteeing that it has no momentum.  In that case, you want the wavefunction to be as flat as possible. Ideally, it's the same value everywhere.  But then the position of the electron isn't defined, since there's no maximal location and the probability distribution extends over all space: It could be anywhere, but wherever it is, it's definitely not moving.

*In practice, you're always going to get some compromise between these extremes.  Maybe the particle is definitely inside the detector, but the detector's large so the momentum can be measured accurately.  Maybe you know the particle hit a thing with a well-defined location at a specific time, but you're not sure where it went after that or how fast it was moving.  Heisenberg Uncertainty principle says there will always be a trade-off of some sort, and to get rid of that, you don't need to merely be creative in your use of electromagnetic fields, you need to understand the actual mathematics of quantum mechanics well enough to break the above model... or disprove the currently-accepted quantum mechanical understanding of "position", "momentum" and "uncertainty".

A: I think the answer is yes. You can proceed in the following way:

*

*Direct an electron gun in some direction, say Z.

*At some large distance on Z place a detector, say a fluorescent screen.

*Set some speed for the electrons, say about 300 m/s.

*Go inside a rocket and make it move along Z with 300 m/s.

*When the rocket passes by the electron gun, make it shoot some electrons.

*When the rocket arrives at the detector look for a signal. If you see such signal, you can deduce that the electron causing that signal was stationary relative to your rocket for the time between emission and detection. You can also calculate its exact position relative to your rocket.

The Heisenberg uncertainty limits your predictions, it does not limits what exists. In this case, for that specific electron, you can know both its position and momentum with any accuracy. As Heisenberg himself said, this principle does not apply to the past.
