Number of continuously differentiable function $f:R\to R$ satisfying $f\left(x\right)>0$and$f\text{'}\left(x\right)=f\left(f\left(x\right)\right)$ $\mathrm{\forall }\mathrm{x}\in \mathrm{R}$ is/are

Assume that such a function exists
$$\begin{gathered} \begin{array}{l}\therefore {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right)>0\Rightarrow \mathrm{f}\uparrow \\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right)>\mathrm{f}\left(0\right)\mathrm{\forall }\mathrm{x}\in \mathrm{R} \\ {\int }_{\mathrm{x}}^0 {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\mathrm{dx}>{\int }_{\mathrm{x}}^0 \left(0\right)\mathrm{dx};\mathrm{x}<0\end{array} \end{gathered}$$
$\Rightarrow$f(x) < (x + 1) f(0) clearly for $\mathrm{x}\le -1,\mathrm{f}\left(\mathrm{x}\right)\le 0$ contradicting the hypothesis that f (x) > 0 $\mathrm{\forall }$x No such function exist