Is a PRG concatenation using the same input still a PRG? Suppose we have a secure PRG $G: \{0,1\}^s→ \{0,1\}^n$ and we define a $G'(k)$ as:

$$
G'(k) = G(k) \| G(k)
$$

So if $G(k)=\mathbf{i}$, $G'(k) = \mathbf{i} \| \mathbf{i}$

Is $G'$ still a secure PRG?

Based on my own understanding it should be, because the statistical distinguishers will produce the same analogy of zeros and ones, therefore $G'$ should be a secure PRG too.

Is there any other principle about PRGs that $G'$ does not satisfy?
Refering to THIS post

If we call $U_k$ the random variable uniformly distributed over bit strings of length $k$, then a function $g: \{0,1\}^k \to \{0,1\}^m$ is called pseudo-random generator if no feasible(poly-time if you want) algorithm can distinguish $g(U_k)$ and $U_m$ with non-negligible probability.

More formally let $U'_m = g(U_k)$ then the distinguishing advantage of any distinguisher efficient $D$ that we denote as $\Delta^D(U'_m, U_m) = \Pr^{DU_m}[Z = 1] - \Pr^{DU'_m}[Z = 1]$ is negligible.  

Here $Z$ is the output of the distinguisher, and negligible is any suitable notion of "really small"; same for efficient.

Now my ability to distinguish between $U_k$ from your PRNG $g: \{0,1\}^k \to \{0,1\}^m \mathbin\| \{0,1\}^m$ is obviously more than negligible. Because any OUTPUT from your generator has symmetry and if I see this symmetry the generator being used is most probably your generator.