Is angular momentum conserved in all possible axis of rotation (give no external torque)?

So circular motion works out well if everything references the center of the circle. If the momentum of the mass is $vec p$, at radius $vec r$ then:

$$ vec L = vec r times vec p$$

is a constant of motion:

$$dot{vec L} = (dot{vec r}timesvec p) + (vec r times dot{vec p}) = 0 + 0.$$

The 1st one is zero because both $dot{vec r}$ and $vec p$ are proportional to $vec v$; the 2nd is zero because the radius is parallel to the force $dot{vec p}$.

The whole thing goes "wrong" when the central point is not $vec 0$, the center of the circle. A qualitative analysis indicates the angular momentum oscillates (if the origin is moved to a point on the circle, it must be zero at that point)...what gives?

If we move it to $vec a$:

$$ vec L = (vec r - vec a) times vec p $$

$$ dot{vec L} = frac{d}{dt}[(vec r - vec a)times vec p]=-frac{d}{dt}[vec a times vec p]$$

Since $vec a$ is fixed:

$$ dot{vec L} = vec a times dot{vec p} = vec{tau}$$

So yes, there is a coordinate dependent oscillatory amplitude (and bias) added to what we would nominally call "the angular momentum", but at least the oscillatory part is explained by a coordinate dependent torque that conforms to our definition of torque and rate of change of angular momentum.

In summary: the radial centripetal force applies a coordinate dependent torque that accounts for the position dependent angular momentum.

In some circles, the fact that these quantities depend on choice of origin is why they are called "pseudo-vectors": they rotate like vectors by don't translate like them. This is related to the fact that they are axial-vectors, which rotate like vector but under reflection, do not change sign like true vectors... and that, is because they're really the 3 antisymmetric components of a rank-2 tensor (so when we say "$z$ component of an (axial)vector", we really mean "$xy$ component of a tensor minus the $yx$ component").