I am assuming that $n$ has a primitive root otherwise the problem makes not sense.Assume that $g$ is a primitive root modulo $n$ then all of the numbers $g,g^3,\dots g^{\varphi (n) -1}$ are quadric residues.So any odd number between $1$ and $\varphi (n)$ must be relativly prime to $\varphi (n)$ which means $\varphi (n)$ has no prime factors which are not Fermat primes or $2$.SInce $2^{2^6}+1$ has more than $20$ digits.And between first $5$ Fermat numbers only the fifth is not a prime the solutions are: $2,4,F_1,F_2,F_3,F_4,2F_1,2F_2,2F_3,2F_4$ Where $F_i$ denotes the $i$'th Fermat number.

Taha1381 wrote: I am assuming that $n$ has a primitive root otherwise the problem makes not sense.Assume that $g$ is a primitive root modulo $n$ then all of the numbers $g,g^3,\dots g^{\varphi (n) -1}$ are quadric residues.So any odd number between $1$ and $\varphi (n)$ must be relativly prime to $\varphi (n)$ which means $\varphi (n)$ has no prime factors which are not Fermat primes or $2$.SInce $2^{2^6}+1$ has more than $20$ digits.And between first $5$ Fermat numbers only the fifth is not a prime the solutions are: $2,4,F_1,F_2,F_3,F_4,2F_1,2F_2,2F_3,2F_4$ Where $F_i$ denotes the $i$'th Fermat number. I agree with Fermat numbers, but in the case of $n=4$, 2 is a quadratic non-residue but not a primitive root of 4.

$n$ has a primitive root when $n$ is 2, 4, $p^{k}$, or $2p^{k}$ for odd primes $p$ and integers $k$. 1) We first consider $n=2$ and $n=4$. It can be easily seen that $n=2$ works, but that $n=4$ doesn't. 2) Now, we consider when $n = p^{k}$. If $k \geq 2$, we show that $n$ doesn't work. This is because $p$ is a quadratic non-residue of $ \pmod{p^k}$ but not a primitive root of $\pmod{p^k}$. So, for this case to work we need $k=1$. Now, we count the number of quadratic non-residues and primitive roots. The number of quadratic non-residues is half of all numbers less than p, so $\frac{p-1}{2}$, and the number of primitive roots is $\phi(\phi(p)) = \phi(p-1)$. Since all quadratic non-residues should be primitive roots, the number of primitive roots should be greater than that of quadratic non-residues. So, $\frac{p-1}{2} \leq \phi(p-1)$ and this is only possible when the prime divisor of $p-1$ is only 2, which yields Fermat numbers. 3) Same argument from 2) can be applied. There are no solutions in this case.