This question expands on this one and seems to have a stronger result.

Take the Riemann $\xi$-function $\xi(s) =\frac12 s\,(s-1) \,\pi^{-\frac{s}{2}}\, \Gamma\left(\frac{s}{2}\right)\, \zeta(s)$. We know that $\xi(s)$ has the same non-trivial zeros as $\zeta(s)$ and also that $\xi(s) = \xi(1-s)$.

With $a \in \mathbb{R}/0$, "stretch" $\xi(s)$ as follows:

$$f(s,a):= \xi(a\,s) \pm \xi\left(a\,(1-s)\right)$$

I like to conjecture that for each $a$, *all* zeros (in and outside the strip) reside on the critical line.

The graph below shows the first zeros ($\pm = +$), all at $\Re\left(\frac12\right)$, for $a$ from $-2$ to $2$ in steps of $0.01$.

At first sight this "curtain" graph looks quite symmetrical, but clearly isn't so.

**Question:**

1) Are there any exceptions that contradict this conjecture?

2) If not, is this just a consequence of the RH or would the RH be a consequence of the conjecture?

3) Since $\xi(s)$ is an entire function that can be expressed as a Hadamard product involving its zeros as factors, $f(s,a)$ should have a unique one too. Appreciate any ideas around what it could look like.

*Note:*

I believe the same conjecture could be applied to:

$$\xi(a+s) \pm \xi\left(a + 1-s\right), \text{ all zeros on } \Re(s)=\frac12$$

and

$$\xi(a+s) \pm \xi\left(a - s\right), \text{ all zeros on } \Re(s)=0$$

Both are symmetrical around $a=0$ and $a=\frac12$ respectively, also the values where the $\rho$'s reside.