$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse.
Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the $z$-direction $\ket{\uparrow_z}$,$\ket{\downarrow_z}$.
Now, the state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{\uparrow_z} + \ket{\downarrow_z})$ is not an eigenstate of $S_z$, and measurement of $S_z$ will collapse it with equal probability into $\ket{\uparrow_z}$ or $\ket{\downarrow_z}$. Yet, this is an eigenstate of the spin in $y$-direction, i.e. $\ket{\psi} = \ket{\uparrow_y}$ (or down, we'd have to check that by computation, but it doesn't matter for this argument). So, although you can "collapse" $\ket{\psi}$ into other states, it already looks like a collapsed state by your logic. This shows that the notion of "looking like a collapsed state" is not very useful to begin with.
Furthermore, you seem to be confused about the difference between a measurement (inducing collapse in some dictions) and the application of an operator. You say
So when the hamiltonian acts on a system in one of those eigenfunctions, that system always collapses to the same point in space?
but this is non-sensical. The action of the Hamiltonian is an infinitesimal time step, as the Schrödinger equation tells you:
$$ \mathrm{i}\hbar\partial_t \ket{\psi} = H\ket{\psi} $$
and, for an eigenstate $\ket{\psi_n}$, which is a solution to the time-independent equation with energy $E_n$, you have by definition $H\ket{\psi_n} = E_n\ket{\psi_n}$, that is, the Hamiltonian is a "do nothing" operation on eigenstates since multiplication by a number does not change the quantum state. That is, after all, why we are interested in the solutions to the time-independent equation - because these are the stationary states that do not evolve in time. This has nothing to do with collapse, or measurement.
Lastly, exactly determinate states of position are not, strictly speaking, quantum states, since the "eigenfunctions" of the position operator "multiplication by x" are Dirac deltas $\psi(x) = \delta(x-x_0)$, which are not proper square-integrable functions $L^2(\mathbb{R})$ as quantum states are usually required to be. But yes, this is "a spike", and conversely, the determinate momentum states are plane waves $\psi(x) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar}px}$.
Best Answer
The position collapses into a position eigenstate (a spike in wavefunction) because you measured the position. That spike spreads over time because the position eigenstates are not momentum eigenstates, i.e. there is a range of momenta in the spike. Another way of looking at this is that the spike is not an eigenstate of the Hamiltonian (i.e. a well defined energy) and therefore evolves in time beyond simple a phase change.
If you measure the energy, you will collapse into an energy eigenstate, which is not a position eigenstate. For example, energy eigenstates of a particle in a box (classic quantum homework problem) are distributed across the width of the box, not uniformly, but not strongly localized like the spike. If you measure the energy again, you will get the same answer (it is an eigenstate of the Hamiltonian).
As for measurements, we measure energy differences usually. For example, the energy eigenstates of a hydrogen atom (or hydrogen like atom) correspond to various configurations of the electron "orbiting" the atom plus relative orientation of the electron's magnetic moment (spin) to the "orbit" and the nuclear magnetic moment.
I work with calcium ions that effective have a single electron around a nucleus (the other electrons effectively cancel out). If the calcium ion is in the lowest energy configuration (ground state) and I shine a laser on it, sweeping the frequency of the laser, the ion will remain dark until my laser is in a narrow ~20 MHz window around 755 THz (397 nm UV light) at which point the ion will glow ... something we can see with a sensitive CCD camera. That 755 THz corresponds to the energy difference between two different configuration of the ion. The ion glows because it can absorb the light at that specific frequency and will then re-emit the light into a random direction through spontaneous emission. The absorption and emission happen about 20 million times a second. If the ion is not in the ground state, and there are other metastables levels it could be in, the ion won't glow. In this way I can detect whether it was in the ground state. When I turn off the laser, the ion will emit a single final photon and fall back into the ground state (and you can detect this photon).
If the ion was originally in some superposition of the ground state and one of the metastable states, the ion will either collapse into the ground state and glow or collapse into the metastable state and stay dark. The probability of it being bright or dark depends on the original superposition. Through this measurement process, I have collapsed the ion's state into an energy eigenstate. At no point are we measuring the electron's position or momentum, only the energy difference.
The following link about last year's Nobel prize might help.