Collapse of the wavefunction after measurement

As stated in the comment made by @anna v, you can't perform a second measurement instantly. That would be the same as performing one measurement.
In the case of position measurement, the wavefunction will have evolved already when you make a second measurement. When making the second measurement a very short time after the first, the wavefunction has already evolved into a continuous set of position eigenstates.
Even when $Delta t$ approaches $dt$ this will be the case. The set of eigenstates, in this case, will have risen too, but with an amount approaching zero.

What about other systems? The same picture. All quantum mechanical states evolve in the same way in time.
Take for example the case of the superposition of two states. After you have done the first measurement (resulting in one out of two eigenstates), the wavefunction will instantly start to develop into a new superposition of two eigenstates. And again it depends on how long after the first measurement you perform the second one.

$|Psi (0)rangle$ is not necessarily an eigenstate of H , however, because H is time-independent, the evolution of $|psi rangle$ is

$displaystyle |{Psi (t)}rangle = e^{-iHt/hbar } |{Psi (0)}rangle = Big(1 + frac{-iHt}{hbar } + frac{1}{2}left(frac{-iHt}{hbar }right)^2 + cdots Big) |{Psi (0)}rangle , .$
Thus,

$displaystyle begin{split} langle {Psi (0)} | {Psi (t)} rangle & = 1 + frac{-it}{hbar } langle {Psi (0)}H|{Psi (0)}rangle + frac{-t^2}{2hbar ^2}langle {Psi (0)}H^2|{Psi (0)}rangle + O(t^3) & = 1 + frac{-it}{hbar } langle Hrangle + frac{-t^2}{2hbar ^2}langle H^2rangle + O(t^3) & = 1 + frac{-it}{hbar } langle Hrangle + frac{-t^2}{2hbar ^2}Big((Delta H)^2 + langle H rangle ^2Big) + O(t^3) end{split}$
where expectation values $langle H rangle$ and $langle H^2 rangle$ are taken with respect to $| psi (0) rangle$, and in the last step I have used that $(Delta H)^2=langle H^2 rangle−langle H rangle^2 $.

Thus, to order $t^2$,

$displaystyle begin{split} big | langle {Psi (0)}|{Psi (t)}rangle big |^2 & = Big[1 + frac{-it}{hbar } langle Hrangle + frac{-t^2}{2hbar ^2}Big((Delta H)^2 + langle H rangle ^2Big)Big] & qquad qquad Big[1 + frac{it}{hbar } langle Hrangle + frac{-t^2}{2hbar ^2}Big((Delta H)^2 + langle H rangle ^2Big)Big] + O(t^3) & = 1 + frac{-t^2}{hbar ^2}Big((Delta H)^2 + langle H rangle ^2Big) + frac{t^2}{hbar ^2}langle H rangle ^2 + O(t^3) & = 1 - frac{t^2}{hbar ^2}(Delta H)^2 + O(t^3) , . end{split}$

As an interesting side point, if t is very small then $big | langle {Psi (0)}|{Psi (t)}rangle big |^2 approx 1$ (to order t ). In other words, if we measure whether $langle psi (t) rangle$ is in the state $langle psi (0) rangle$ after only letting the system evolve for a very short time, we expect with high probability to indeed find an outcome of $| psi (0) rangle$, i.e. the system effectively hasn't evolved at all! If, after letting the system evolve for another very short time, we measure again, we again likely expect to find the state to be $| psi (0) rangle$. This is the source of the so-called quantum Zeno effect; if you continually measure a quantum state, it does not change, or colloquially, a watched quantum pot never boils