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**This is not a complete answer, but only an attempt for it.**

It should not be surprising that considering the specific form $f(x)=k\sin(jx+y)$ , the condition $f'(0)=1$ fixes value of all the 3 paramters $k,j,y$:
* Since $1$ is the maximum value $f'(x)$ is allowed to attain, it must be that $y=0$ (other other value resulting in identical function $f$ due to its periodicity).
* The condition $jk=1$ together with $|k|\leq 1$ and implies that $k=j=1$. Indeed, otherwise $|j|>1$, and so $f^{(n)}(0)$ would be greater than $1$ in absolute value for some $n$ large. 

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What I actually find surprising is how hard it seems to find any function $f$ with $f'(x)=1$ and $|f^{(n)}(x)|\leq 1$ for all $n=0,1,\ldots$ and $x\in \Bbb R$. A possible approach to do so is to consider the functions in the form
$$
  f(x) = \sin(g(x)),
$$
where $g(x)$ is some function with $g'(0)=1$. Then the derivatives of $f$ are
\begin{align}
  f'(x) &= g'(x) \sin(g(x)) \\
  f''(x) &= g''(x) \sin(g(x)) + [g'(x)]^2 \sin(g(x)) \\
  f^{(3)}(x) &= g^{(3)}(x) \sin(g(x)) + 2g''(x)g'(x)^2 \sin(g(x))+ [g'(x)]^3 \sin(g(x)) \\
&= \left[\left(\frac{d}{dx}+g'(x)\right)^2 g'(x) \right] \sin(g(x)),\\
 f^{(n)}(x) &= \left[\left(\frac{d}{dx}+g'(x)\right)^{n-1} g'(x) \right] \sin(g(x)),
\end{align}
where $\left(\frac{d}{dx}+g'(x)\right)^{n-1}$ is first expanded using  the [binomial formula](https://en.wikipedia.org/wiki/Binomial_theorem), and finally applied to $f'(x)$ in the sense that $\left(\frac d {dx}\right)^n f'(x) = f^{(n+1)}(x)$.

Unfortunately, I don't know how to continue from here. I tried to consider functions $g$ with $g'(0)=1$ and whose derivatives would be vanishing as $|x|\to \infty$, but there is always some derivative that happens to be greater than 1.