What can be said about a function that is odd (or even) with respect to two distinct points? This question is a little open-ended, but suppose $f : \mathbb R \to \mathbb R$ is odd with respect to two points; i.e. there exist $x_0$ and $x_1$ (and for simplicity, let's take $x_0 = 0$) such that
$$
(1): \quad f(x) = -f(-x)
$$
and
$$
(2): \quad f(x_1 + x) = -f(x_1 - x)
$$
for all $x$.
Then, the vague question I'd like to answer is

What else can we conclude about this function?

It seems maybe I can conclude it is periodic, with period $2x_1$ (in general $2\left|x_1 - x_0\right|$): given the values of $f$ on $(0,x_1)$, $(1)$ determines the values on $(-x_1,0)$, then this and $(2)$ determine the values on $(x_1,3x_1)$, then this and $(1)$ determine the values of $f$ on $(-3x_1,-x_1)$, this and $(2)$ determine the values on $(3x_1,5x_1)$, and so forth.
Is there anything else to say here?
 A: You can conclude that it's periodic with period dividing $2x_1$, for essentially the reason you gave.
On the other hand, if $f$ is odd (about zero) and periodic with period dividing $2x_1$, then
$$
f(x_1+x)=f(2x_1-x_1+x)=f(-x_1+x)=-f(x_1-x)
$$
and so $f$ is also odd about $x_1$. So "odd about two points" is equivalent to "odd and periodic."
A: No. There's really not much else to say. You get a periodic function that's got a certain symmetry (on some period, it's "even", and on a period offset from this by a half-period, it's also even) from your conditions. But if I give you a periodic function satisfying this "double evenness" property on some period, then it'll turn out to be doubly-even as a function on the reals. So the two ideas are really one and the same. 
The double-evenness probably implies something about the fourier coefficients, esp. if one of the "symmetry points" is the origin. I guess that even single-evenness tells you that all the "sine" coefficients are zero. The other evenness probably says something like all the odd  (or all the even?) cosine coefficients are also zero, but I haven't worked out the details. 
