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:
$$
\sigma_x \sigma_p \ge \hbar/2
$$
