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For $c=0$ the following should be true for the $n$-th Gram point:
$$frac{x}{2 pi e}logleft(frac{x}{2 pi e}right) = frac{x}{2 pi e}log left(frac{x}{2 pi e}right) + frac{-c+n}{e}-frac{vartheta (x)}{pi e}$$
and for $c=frac{1}{2}$ it should be true for the $n$-th complementary Gram point. $vartheta (x)$ is the Riemann-Siegel theta function.

That equation can be solved for $x$ like this:

$$x=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$

where $W$ is the Lambert W function.

Then name the right hand side the function $f(x)$:
$$f(x)=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$

and iterate it as many times as computer power allows. This iteration is the normalizing function that we call $g(n,x)$:

$$g(n,x)=f(...f(f(f(f(x))))...)$$

This should allow us to cut the real part of the Riemann zeta function:

$$Releft(zeta left(frac{1}{2}+i g(n,1)right)right)$$

at complementary Gram points $c=frac{1}{2}$ into a family of functions with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $x=1$.
In case I did not say this correctly, this how it is written in Mathematica:

Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]

Plot of the superimposed real parts:

where the end points at $0$ and $1$ are complementary Gram points.

The average of the 101 curves looks like this:

This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:

Mathematica code for the normalized real parts:

Clear[x, n, c, f, g]

c = 1/2;

f[x_] = 2*Pi*E*

   E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n - 

        RiemannSiegelTheta[N[x]]/Pi)/E];

g[y_] = f[f[f[f[y]]]];

Show[Monitor[

  Table[ListLinePlot[

    Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], 

    DataRange -> {0, 1}, PlotRange -> {-1.3, 6.4}], {k, 0, 100}], k]]

Clear[x, n, c, f, g]

c = 1/2;

f[x_] = 2*Pi*E^1*

   E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n - 

        RiemannSiegelTheta[x]/Pi)/E^1];

g[y_] = f[f[f[f[y]]]];

Show[Monitor[

  ListLinePlot[

   Sum[Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0, 

      100}]/101, DataRange -> {0, 1}], k]]

We can do the same for the imaginary part of Riemann zeta:

$$Imleft(zeta left(frac{1}{2}+i g(n,1)right)right)$$

Again we look at the family of functions between complementary Gram points $c=frac{1}{2}$ with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $y=1$.
In Mathematica:

Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]

The plot of the superimposed imaginary parts looks like this:

where the middle common point at $frac{1}{2}$ is at a complementary Gram point.

The average of the 101 curves is:

This again looks like $sin (x)$ on the interval $left[0,2piright]$:

Would such a normalization with the function $g(n,x)$, be any easier to study, than the Riemann zeta function directly?

Mathematica code for the normalized imaginary parts:

Clear[x, n, c, f, g]

c = 1/2;

f[x_] = 2*Pi*E*

   E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n - 

        RiemannSiegelTheta[N[x]]/Pi)/E];

g[y_] = f[f[f[f[y]]]];

Show[Monitor[

  Table[ListLinePlot[

    Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], 

    DataRange -> {0, 1}, PlotRange -> {-5, 5}], {k, 0, 100}], k]]

Clear[x, n, c, f, g]

c = 1/2;

f[x_] = 2*Pi*E^1*

   E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n - 

        RiemannSiegelTheta[x]/Pi)/E^1];

g[y_] = f[f[f[f[y]]]];

Show[Monitor[

  ListLinePlot[

   Sum[Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0, 

      100}]/101, DataRange -> {0, 1}], k]]

I don't know what you want to say with this iteration.

See this. Defining the arguments correctly (zero on $s > 1$, continuous on $Re(s) > 1$, and extended by continuous variation on horizontal lines to $Re(s) le 1$) we can express the number of non-trivial zeros $zeta(rho) = 0, Im(rho) in (0,t]$ with $$ N(t) =frac{text{arg }(Z(t))}{pi} = frac{vartheta(t)}{pi}+ frac{text{arg }(zeta(1/2+it))}{pi} = frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t) + mathcal{O}(log t)$$ where $frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t)$ comes from the Stirling approximation applied to $vartheta(t)$, and $ mathcal{O}(log t)$ comes from a bound for $text{arg }(zeta(1/2+it))$.

Assuming the Riemann hypothesis and that $text{arg }(zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$rho_n approx 1/2+i vartheta^{-1}(n)$$