If nG and 2nG are known how to find generator point G If in a curve $y^2 = x^3+ 7 \mod p$ ( ecdsa-secp256k1)

p=115792089237316195423570985008687907853269984665640564039457584007908834671663

G=55066263022277343669578718895168534326250603453777594175500187360389116729240,32670510020758816978083085130507043184471273380659243275938904335757337482424

nG=21505829891763648114329055987619236494102133314575206970830385799158076338148,98003708678762621233683240503080860129026887322874138805529884920309963580118

2nG=72488970228380509287422715226575535698893157273063074627791787432852706183111,62070622898698443831883535403436258712770888294397026493185421712108624767191

How do I find $n$ ?
You are asking two different questions in the title and the body. The answer to both questions is "You can't", but for different reasons.

Finding $n$:

You're asking how to find $n$, given $X=n\cdot G$ and $2\cdot X=2n\cdot G$, where $G$ is the generator of some group $\mathbb{G}$ and $n$ is sampled uniformly.
We first note, that being given $X$ and $2\cdot X$ is equivalent to just being given $X$. We could just compute $2\cdot X$ ourselves. Consequently, finding $n$ is equivalent to solving the discrete logarithm problem in $\mathbb{G}$. As the discrete logarithm problem is assumed to be hard in the used group, that's not going to happen.

Finding $G$:

You are asking how to find the generator $G$ of group $\mathbb{G}$, given $n\cdot G$ and $2n \cdot G$. This is impossible in general. Most groups have more than one generator, the group you propose to use is no exception to that. Given two generators $G,G'$ there by definition exists an $x$ such that $G = x\cdot G'$. Therefore, $n \cdot G = nx \cdot G'=n'\cdot G'$ and $2n \cdot G = 2nx \cdot G'=2n'\cdot G'$ and given no further information about $n$, it is impossible to decide which of the two generators was originally used.