Is there any characterization of continuous functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ such that for any linearly independent set $A$ (over the rationals) $f(A)$ is also linearly independent ? 



Continuous functions $f$ with $f(A)$ linearly independent when $A$ is independent
This may not be a complete answer to the question, but it is too long for a comment.


Given such a function $f$, the restriction $f \restriction [0,\infty)$ is a power function $\alpha x^\beta$ where $\alpha \ne 0$ and $\beta \ge 0$, and similarly the restriction $f \restriction (-\infty,0]$ is a power function of $-x$. We will prove this for the restriction $f \restriction [0,\infty)$ because the other case is similar.


Because $\sqrt{2}$ is irrational, for any $x > 0$ the set $\{x,\sqrt{2} x\}$ is linearly independent, so the set $\{f(x),f(\sqrt{2} x)\}$ is linearly independent, meaning that its elements are nonzero and either


1. Their quotient $f(\sqrt{2} x)/f(x)$ is irrational, or
2. They are the same element: $f(\sqrt{2} x) = f(x)$, so the quotient $f(\sqrt{2} x)/f(x)$ is 1.


In either case the quotient $f(\sqrt{2} x)/f(x)$ must take a constant value on this interval because it is a continuous function of $x$ whose range is contained in the totally disconnected set $(\mathbb{R} \setminus \mathbb{Q}) \cup \{1\}$.


Therefore the function $x \mapsto f(2 x)/f(x)$ must also take a constant value $K$ on the interval $(0,\infty)$, namely the square of the previous constant.


Observe that for any positive integer $n$ we have $f(2^{1/n}) = K^{1/n}f(1)$, and in fact for any positive rational number $q$ we have $f(2^q) = K^q f(1)$.
This means that $f$ is a multiple of a power function on the set $\{2^q : q \in \mathbb{Q}\}$. By continuity this holds on the closure of this set, namely the interval $[0,\infty)$.


Note that the constant factor $\alpha$ cannot be zero, and that the exponent $\beta$ must be non-negative for $f$ to be continuous at zero.


*Discussion of special cases:*


Constant case: If the exponent on the interval $[0,\infty)$ is zero, then
the function is a nonzero constant $\alpha$ on this interval.
Therefore $f(0) \ne 0$, which means that on the interval $(-\infty,0]$ the exponent must also be zero and the function must have the same constant value $\alpha$ by continuity. So in this case $f$ is a nonzero constant function.


Another case is that the exponent $\beta$ is equal to $1$ on both the intervals $(-\infty,0]$ and $[0,\infty)$, so it is piecewise linear.
In this case the only restriction on the slopes is that they must be nonzero rational multiples of each other. I don't know if this is the only other case.


