Remove the order $1$ pole of $\zeta(s)$ at $s=1$, to create the following entire function:
$$z(s):=\zeta(s)-\dfrac{1}{s-1}$$
I like to conjecture that all complex zeros of $z(s) \pm z(1-s)$ in the critical strip lie on the line $\Re(s)=\frac12$. Note that there are a few complex zeros lying outside the strip, however these seem 'capped' at |$\Im(s)| < 6$ and also rapidly converge towards zeros for real values only at $s=\frac12 \pm (2k + \frac12)$ with $k \in \mathbb{N}, k>8$.
I include a root finding graph below for $z(s)^2 - z(1-s)^2$ (to induce the roots for both the sum and the difference from a single function) over the domain $\Re(s)=-27..27$ and $\Im(s)=0..33$ :
Would this result be a consequence of RH being true ?
EDIT:
I was too quick. Now found a few zeros lying off the critical line at higher $\Im(s)$ (e.g. $0.4793654900+260.7864439 *i*$ , so I clearly have to withdraw my claim.