What is function $f,g:\mathbb R^+\to\mathbb R$ sought that satisfies $$\forall x\in\mathbb N,\,f_{(r)}(x)=\underbrace{f(f(\dots(f(f(x)))\dots))}_{r\text{ times}}=2^{(\log x)^c}$$ $$\forall x\in\mathbb N,\,g_{(r)}(x)=\underbrace{g(g(\dots(g(g(x)))\dots))}_{r\text{ times}}=2^{x^\frac{1}{c}}$$ where $c>0$ is fixed?
What is solution if above relations hold at every $x\in\mathbb R$?
With $$ u(x) = \left\{ \begin{array}{cl} 1 & {\small\mbox{for }} x = 0 \\ 0 & {\small\mbox{for }} x \ne 0 \end{array} \right. $$
We define \begin{align} f(x) &= \left[\sum_{k=1}^\infty 2^{(\log k)^c} u(x-k) \right] + x \left[ 1 - \sum_{k=1}^\infty u(x-k)\right] \\ &= x + \sum_{k=1}^\infty \left(2^{(\log k)^c} - x\right) u(x-k) \end{align}
then for $n \in\mathbb{N}$ $$ f(n) = 2^{(\log n)^c} \quad (*) $$
and for $x \not\in\mathbb{N}$ $$ f(x) = x $$ For $n \in \mathbb{N}$ this gives $$ f_{(r)}(n) = f^{r}(n) = f^{r-1}(f(n)) = f^{r-1}(2^{(\log n)^c}) = 2^{(\log n)^c} $$ The construction relies mostly ($n=1$ gives $f(1) = 1$) on the assumption that natural numbers do not get mapped to natural numbers in equation $(*)$.
The shorter version would be $$ f(x) = \left\{ \begin{array}{cl} 2^{(\log x)^c} & {\small\mbox{for }} x \in\mathbb{N} \\ x & {\small\mbox{for }} x \not\in\mathbb{N} \end{array} \right. $$
