Is the result $\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1\implies \Re(s)=\frac{1}{2}$ already known / meaningful towards the RH? As an undergrad, I sometimes toy with the Riemann Hypothesis (RH) to A) humble myself and B) better myself in complex analysis and proof writing. Given its infamy and my relative mathematical immaturity, I obviously don't expect to make any significant results any time soon. Hence, I apologize for fulfilling the often annoying trope of "young ambitious undergrad attempting millennium prize problems."
That said, after some playing around with the zeta function for the past few years, I believe I have come across a proof that
$$\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1\implies \Re(s)=\frac{1}{2}$$
for all $s$ within the critical strip.

Assuming this is true, I am curious if this result is meaningful to RH. Is this result already known, or a triviality of the functional identity? If not, is it meaningful / significant to the RH?

Any insight as to its potential (non)significance, how this may be false, or how this is already known would be greatly appreciated.
Edit: As pointed out, this 'result' (or conjecture, if you will) fails near the origin. I'll have to overlook my proof again, but fortunately I believe it should still hold for $|\Im(s)|\ggg 0$, at the least.
 A: Intro: I am sorry for the following lines since your result would have been an outstanding achievement towards RH.However... 
By the reflection formula
$$ \rho(s)=\frac{\zeta(s)}{\zeta(1-s)} = 2^s \pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s) $$
and
$$ R(\nu)=\rho\left(\frac{1}{2}+\nu\right) =(2\pi)^{\nu} \sqrt{\frac{2}{\pi}}\sin\left(\frac{\pi \nu}{2}+\frac{\pi}{4}\right)\Gamma\left(\frac{1}{2}-\nu\right)$$
is a real analytic function on $(-1/2,1/2)$ which is also increasing on such interval and equal to $1$ at the origin. For any $\nu\in(-1/2,1/2)$
$$\|R(\nu+i\tau)\|^2 = R(\nu+i\tau)R(\nu-i\tau) = (2\pi)^{2\nu}\left(1+\frac{\sin(\pi\nu)}{\cosh(\pi \tau)}\right)\left\|\frac{\Gamma\left(\tfrac{1}{2}-\nu-i\tau\right)}{\Gamma\left(\tfrac{1}{2}-i\tau\right)}\right\|^2$$
is an even function of the $\tau$ variable. Assuming $\nu > 0$ it is also a Schwartz function.
On the other hand, by just considering some concrete instances, we may realize that your conjecture does not hold. For $\nu=\frac{1}{10}$ we have $\|R(\nu)\|^2>1.72$ and $\|R(\nu+i\tau)\|=1$ for $|\tau|\approx 6.29 $. In particular there are points on the line $\text{Re}(s)=\frac{3}{5}$ such that $|\zeta(s)|=|\zeta(1-s)|$. But exactly two of them.

On a positive note, let us try to salvage things. 
Even if your conjecture does not hold, it leaves some room for an interesting attack. Let us set $s=\sigma+i\tau$ for $\sigma\in (0,1)$.
Jack's conjecture 1: The locus of points of the critical strip such that $|\zeta(s)|=|\zeta(1-s)|$ is made by the line $\sigma=\frac{1}{2}$ and by the union of two graphs, symmetric with respect to the real line, given by $\tau=\pm f(\sigma)$.
Jack's conjecture 2: In order to prove RH it is sufficient to show that there are no zeroes of the $\zeta$ function on the curve $\tau=f(\sigma)$.
Numerically it seems that $f$ is almost constant, taking values between $6.28$ and $6.3$. 
And the non-trivial zero closest to the origin is well-known to fulfill $\tau > 10$ (this just follows from integral representations for $\zeta(s)$).
If this actually works we'll share the prize, deal? :)

Update. As shown in the comments, there is no hope to prove RH through this naive approach, but I am still convinced that something (like improving the shape of the zero-free region) can be done. Let us assume that all the exceptional zeroes in the critical strip are simple. Then for any one of them the limit of $\frac{\zeta(s)}{\zeta(1-s)}$ equals $-\frac{\zeta'(s)}{\zeta'(1-s)}$, so they have to lie on the region defined by the equality between $f(s)=\left|\frac{\zeta(s)}{\zeta(1-s)}\right|$ and $g(s)=\left|\frac{\zeta'(s)}{\zeta'(1-s)}\right|$. $f(s)$ essentially is an elementary function and its behavior for a fixed $\sigma\in\left(\frac{1}{2},1\right)$ and a varying $\tau\in\mathbb{R}$ is Gaussian-like. $\zeta'(s)$ and $\zeta'(1-s)$ are still related via the derivative of the reflection formula, but the ratio $\frac{\zeta'(s)}{\zeta'(1-s)}$ is less elementary, depending on bounds for $|\zeta(s)|$. Numerical experiments reveal that the behaviour of $g(s)$ for a fixed $\sigma$ and a varying $\tau$ still is essentially Gaussian, but for $\tau$ close to zero $g(s)$ is much larger than $f(s)$. If we prove that the curve given by $f(s)=g(s),\tau > 0$ lies in the currently known zero-free region, we have proven RH under the assumption that all the exceptional zeroes are simple.
