What is known about the functional square root of the Riemann zeta function?

Question

Let us consider the Riemann zeta function $\zeta(s)$, where $s$ can take on values on the domain $\mathbb{R}_{>1}$:

$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$

I wonder what is known about the functional square root(s) of the Riemann zeta function defined on the aforementioned domain (*). In other words, I'm curious about the properties of the function(s) $f$ such that $$f(f(s)) = \zeta(s). \qquad \qquad (1)$$

Questions

1. Has a closed-form solution been found for $f$ in equation $(1)$ ?
2. If not (which I expect), have partial results been found for such a function? Properties like existence, (non)uniqueness, continuity, or results about the functional square root of the partial sums? $$f(f(s)) = \sum_{n=1}^{k} \frac{1}{n^{s}} \qquad \qquad k \in \mathbb{Z}_{>0}$$
3. If so, I would be grateful if you have some pointers to relevant articles or other sources.

(Cross-post from MSE.)

(*) Edit as per Gerald Edgar's answer, this condition should be changed. We must define $f$ on $(0, \infty) \cup X$ for some subset $X \subset \mathbb{R} \setminus (0,\infty) $. Then $f$ must map $(1,\infty)$ bijectively onto $X$, and $X$ iself onto $(0,\infty)$. Under these conditions, there is still a possibility that $f$ is both continuous and real-valued. I am interested in the properties of such an $f$.

Answer 1

Note that $\zeta$ maps $(1,\infty)$ bijectively onto itself, $\zeta$ is continuous, $\zeta$ is decreasing on $(1,\infty)$.

Suppose $\zeta = f \circ f$ where $f$ also maps $(1,\infty)$ onto itself. A continuous injective function $f$ from an interval to an interval is either increasing everywhere or decreasing everywhere. (This is from the intermediate value theorem.) But in either case, $f\circ f$ is increasing, so $f \circ f \ne \zeta$.

Now of course we can define $f$ on some larger set, say $(1,\infty) \cup X$, where $X$ has the power of the continuum. Let $f$ map $(1,\infty)$ bijectively onto $X$ and $X$ bijectively onto $(1,\infty)$. Easy.