CSIDH Squaring Fixing the Base Curve Consider the following variants of the CSIDH squaring problem.
P1. Given $sE, E$ where $s$ is a random ideal class and $E$ is a random curve (reachable from initial $E_0$), find $s^2E$
P2. Given $sE_0$ where $s$ is a random ideal class and $E_0:y^2=x^3+x$ is a fixed initial curve, find $s^2E_0$.
Of course, there's also the decisional variant:
DP1. Given $sE,tE,E$ where $E$ is a random curve as above and $s,t$ are either i.i.d. sample of ideal class or $s^2=t$, decide which.
DP2. Given $sE_0,tE_0$ where $E_0:y^2=x^3+x$ and $s,t$ are either i.i.d. sample of ideal class or $s^2=t$, decide which.
What do we know about the hardness of (D)P1 compared to (D)P2? Are they even comparable?
I would also like to stress that, for another equivalent problem called the inverse problem stated in P3, fixing the base curve to $E_0$ would weakens P3 to P4 as easy as finding a quadratic twist. However, computing quadratic twists would only reduce (D)P2 to itself. So it is not so obvious whether the hardness of (D)P2 is weaken.
P3. Given $sE,E$ find $s^{-1}E$.
P4. Given $sE_0$ find $s^{-1}E_0$.
If you have a solver for P1, then it can also solve P2, so the problems are comparable: P2 is easier.
P1 is assumed hard, and is used as the basis of, e.g. https://eprint.iacr.org/2020/1012.
P1 is known to be generically equivalent to P3 (see https://hal.archives-ouvertes.fr/hal-02373179 and https://eprint.iacr.org/2020/1012), however P2 is probably not equivalent to P4: the reduction does not work for these more constrained problems, and P4 is known to be easy, whereas P2 is likely not.