$\newcommand{\ket}[1]{\lvert #1 \rangle}$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
This is a good question, and the answer depends on how mathematically watertight we want to be.
In the standard physicist's presentation of quantum mechanics, we start with a Hilbert space $\mathcal H$ (e.g. the set of square-integrable functions on a line, $L^2(\mathbb R)$). An observable corresponds to a hermitian (really, self-adjoint) operator $A$ on $\mathcal H$, which has the property that $$\langle \psi, A \phi\rangle = \langle A \psi, \phi\rangle$$
Every hermitian operator $A$ has a set of orthonormal eigenstates $\psi_i$ and corresponding eigenvalues $\lambda_i$, such that $A \psi_i = \lambda_i \psi_i$ and $\langle\psi_i,\psi_j\rangle = \delta_{ij}$
Additionally, this set of eigenstates actually spans the space, so given any arbitrary state $\Psi$, we can expand it like this: $$\Psi = \sum_{i=1}^\infty c_i \psi_i$$
The probability of measuring the observable $A$ to have value $\lambda_i$ is now equal to $|c_i|^2$, and if we perform the measurement and get $\lambda_i$ as a result, the state "collapses" into the corresponding $\psi_i$.
When we apply this to the position operator $X$ (which acts by multiplying wavefunctions by the variable $x$), we say that the eigenstates of $X$ are the delta functions centered at all real numbers, because
$$ x \delta(x-x_0) = x_0 \delta(x-x_0)$$
Therefore, if we measure the position of a particle and get a real number $x_0$, the state collapses into the state $\delta(x-x_0)$, which is an infinitely sharp spike centered at $x=x_0$.
Everything I said (other than sweeping the distinction between self-adjoint and hermitian under the rug) is actually mostly accurate up until I mentioned the position operator. The problem is that the formalism described up to that point only applies to operators with discrete spectra.
The position operator has a purely continuous spectrum - namely $\mathbb R$, for a particle on a line. The problem with this is that such operators actually don't have any eigenvalues or eigenstates. Delta functions don't count because they aren't actually states in $\mathcal H$ - they are not square-normalizable functions (or indeed, functions at all).
The notion of a spectrum is more general than that of eigenvalues. For an operator which only has a discrete spectrum, the spectrum is simply equal to its set of eigenvalues. However there are many operators (position, for example) which have purely continuous spectra, and therefore no eigenvalues at all. There are even operators which have a combination of both, like the energy operator in the case of a hydrogen atom. The spectrum of that operator is discrete for negative energies (corresponding to bound states) but continuous for positive energies (corresponding to scattering states).
To understand operators which have continuous spectra, we need the machinery of the spectral theorem, which would take the better part of a lecture course on functional analysis to develop properly. Instead, I'll just directly address your question.
Let $\psi$ be a wave function. The spectrum of $X$ is just the real line, so consider any interval $I\subseteq \mathbb R$. The spectral projector $P_I$ maps an unmeasured state to the state you'd have after you measured the position to be in the interval $I$.
In the case of the position operator, the appropriate projector is just the indicator function: $$ P_I = \mathbb 1_I(x) = \cases{1 & $x\in I$\\0 & $x\notin I$}$$
I'll be concrete. Let's say that the wavefunction of my particular state is a Gaussian, so $\psi(x) = e^{-x^2}$
Now I devise an experiment to measure whether the particle is in the interval $[\frac{1}{2},1]$. If I perform my measurement and find that my particle is in that interval, then the resulting projector is the function
$$ P = \cases{1 & $x\in[\frac{1}{2},1]$\\0 & elsewhere}$$ and my new state would be this:
On the other hand, if I perform my measurement and find that my particle is not in that interval, then the resulting post-measurement state would be this: