Take the following Hurwitz zeta:

$$\zeta_{H}(s,z)$$

with $s=\sigma \pm ti$ and $\displaystyle z=1 \pm \frac{i}{a}$ and $t,a \in \mathbb{R}$.

In the critical strip $0 \lt \sigma \lt 1$, this Hurwitz zeta appears to have zeros $\rho_H$ for combinations of $(s,a)$ for all values of $\sigma$, with the exception of $\sigma = \frac12$. However, contrary to $\zeta(s)$, these zeros do not seem to obey the rule that when $s=\rho_H$ then the function must vanish also at $1-\rho_H$ .

Example:

$$\zeta_{H}(0.8-24.96910...i, 1+ \frac{i}{44.82238...} = 0 \ne \zeta_{H}(0.2+24.96910...i, 1+\frac{i}{44.82238...})$$

$$\zeta_{H}(0.2+25.05217...i, 1+ \frac{i}{44.84243...} = 0 \ne \zeta_{H}(0.8-25.05217...i, 1+\frac{i}{44.84243...})$$

Would a proof for all $\rho_H$ that:

$$\zeta_{H}(\rho_H,z) \ne \zeta_{H}(1- \rho_H,z)$$

for all $\sigma \ne \frac12$ in the critical strip, imply the RH when $\displaystyle \lim_{a \to +\infty}$?

My logic towards a 'yes' to this question, would be that the contradiction between 'all zeros off the critical line cannot be symmetrical for values of $a < \infty$' and 'all non trivial zeros of $\zeta(s)$ must be symmetrical', can only be solved when 'there cannot be zeros of $\zeta(s)$ lying off the critical line'. This would also imply that both functions quite naturally complement each other (i.e. all zeros lie off the critical line versus all zeros are on it). Not sure though about what happens to the properties of both functions at the tilting point when $\displaystyle \lim_{a \to +\infty} \zeta_{H}(s, 1 \pm \frac{i}{a})$ 'morphs' into $\zeta(s)$.