$f(16x)=16f(x) $ and $ f$ is continuous $f: \mathbb{R}\rightarrow \mathbb{R}$ is a continuous function such that $f(16x)=16f(x)$ for every real $x$. 
Should it be $f(x)=ax$? How can I prove that?
 A: Clearly $f(0)=0$ Also, if we have a solution defined for negative $x$ and for positive $x$, we can simply combine them.
Let $h:\mathbb R\to \mathbb R$ be any continuous function with period $1$ - that is $h(x+1)=h(x)$ for all $x$.
Define $$f(x)=xh(\log_{16} x)$$ for $x>0$.
This gives an arbitrary $f:\mathbb R^+\to\mathbb R$ which satisfies $f(16x)=16f(x)$ and $f(x)\to 0$ as $x\to 0$ (the latter because $h$ is bounded.)
It is not hard to prove this gives all such functions.
This is essentially the same answer as several above, but with a cleaner formula that lets us see the relationship between the "obvious" examples $f(x)=ax$ and this general example,  $f(x)=xh(\log_{16} x)$.
The general solution is, given $g,h:\mathbb R\to\mathbb R$ continuous with period $1$ (that is, $g(x+1)=g(x)$ and $h(x+1)=h(x)$ for all $x$) then:
$$f(x)=\begin{cases}xg(\log_{16} x)&\text{if }x>0\\
xh(log_{16} (-x))&\text{if }x<0\\
0&\text{if }x=0.
\end{cases}$$
is an example, and this covers all examples. 
Given an $f$ which satisfies the above rule, define $$g(t) = \frac{f(16^t)}{16^t}$$ for $t\in\mathbb R$. We can easily show that $g(1+t)=g(t)$ that $g$ is continuous, an then, letting $t=\log_{16} x$ for $x>0$ we see that $f(x) = xg(\log_{16} x)$.
Similarly, let $h(t)=\frac{f(-16^t)}{16^{t}}$.
A: No, this is not true. For example, $f(x)=|x|$. However, it will be true if $f(kx)=kf(x)$ for all $k\in\mathbb{Z}$.
A: Let $g,h\colon[1,16]\to\mathbb R$ be two continuous functions with $g(16)=16g(1)$ and $h(16)=16h(1)$, but otherwise arbitrary.
Define
$$ f(x)=\begin{cases}16^kg(u)&\text{if }x>0\text{ with }k:=\lfloor\log_{16}x\rfloor, u:=\frac x{16^k},\\16^kh(u)&\text{if }x<0\text{ with }k:=\lfloor\log_{16}|x|\rfloor, u:=\frac {|x|}{16^k},\\0&\text{if }x=0.\\
\end{cases}$$
Then $f\colon\mathbb R\to \mathbb R$ is continuous(!) and $f(16x)=16f(x)$ for all $x$.
This gives us a lot of examples and in fact all examples (for given $f$, functions $g=f|_{[1,16]}$ and $h=(-f)|_{[1,16]}$ produce $f$ again).
A: There are tons of non-linear solutions for the functional equation, you can even demand them to be smooth everywhere (except at the origin):
$$f(x) =\begin{cases}
x \left(A_0 + \sum_{k=1}^{\infty} A_k \sin(2\pi k \log_{16}|x| + \alpha_k)\right)& x > 0\\
0 & x = 0\\
x \left(B_0 + \sum_{k=1}^{\infty} B_k \sin(2\pi k \log_{16}|x|+ \beta_k)\right)&  x < 0
\end{cases}
$$
where $A_k, B_k, \alpha_k, \beta_k$ are arbitrary constants.
Notice $f(0) = \lim_{n\to\infty}f(16^{-n}) = \lim_{n\to\infty}16^{-n}f(1) = 0$. 
If one futher demand $f$ differentiable at $0$, then for all $x \ne 0$, we have:
$$\frac{f(x)}{x} = \lim_{n\to\infty}\frac{f(16^{-n}x)}{16^{-n}x} = \lim_{h\to 0} \frac{f(h)-f(0)}{h} = f'(0) \implies f(x) = f'(0)x $$
i.e. The only solutions which are differentiable at $x = 0$ are the linear ones.
