I have tried to show that $|f'(0)|< 1$ and that $|f'(1)|<1$. I don't know if it is correct and if it is, how do I show that it has only one in total and the nature of the fixed point? Derivation of formula not required.
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Study $g:x \mapsto f(x)-x$
You have $g(0) = 1$ and $g(1) = \cos({1\over4})e^{-{1\over4}}-1\lt0$ thus by the intermediate value theorem, $g$ has a root in $[0, 1]$, and therefore $\exists x \in [0,1], f(x) = x$
Also note that the derivative $$f'(x) = -\frac{1}{4} e^{-x/4} \left(\sin \left(\frac{x}{4}\right)+\cos \left(\frac{x}{4}\right)\right)$$ clearly satisfies $|f'(x)|\leq 1/2$ for $x\in[0,1]$. This implies that there is only one fixed point and that this fixed point is attractive.
