If $f(x)=f(2x)$, then $f$ is differentiable

Take any continuous function $f:[1,2]\to\mathbb{R}$ with $f(x)=f(2x)$, then extend it to $(0,\infty)$.
Is it differentiable ? Does it remain uniformly continuous near zero?

An example of function satisfying the given properties is $f(x)=\sin (\frac {2 \pi} {\ln 2} \ln (|x|)$. So there are non-constant continuous functions with $f(x)=f(2x)$. The function in this example is not uniformly continuous because it does not extend to a continuous function on $[0,\infty)$. Hence uniform continuity need not be true. To show that $f$ need not be differentiable we can take a non-differentiable continuous function on $[1,2]$ with $f(1)=f(2)$ and the extend it to the intervals $[2^{n}, 2^{n+1}]$ and $[2^{-n-1}, 2^{-n}]$ using the condition $f(x)=f(2x)$. [ This last part is a repetition of the answer by Empy2 and I am just quoting it here for completeness.