If $ind_r(p)$ is coprime with $q-1$, then $p \equiv r^{ind_r(p)}$ is a primitive root $\bmod q$ (as $ind_r(p)$ is invertible $\bmod q-1$). Conversely, given a prime $q$ for which $p$ is a primitive root, we may pick any prime $i$ coprime with $q - 1$ and obtain a primitive root $r$ for which $ind_r(p) = i$ by $r \equiv p^{i^{-1}}$.
We therefore need to prove that all primes $p$ are primitive roots modulo some prime $q > p$.
Thanks to @metamorhpy, I found an answer to my question here. I detail it there.
We will prove that if $a\geq 0$ and $n\geq 0$ are integers such that $\gcd(a,n) =1$, then the equation $x^x \equiv a\pmod{n}$ always has a solution $x\in\mathbb{N}$.
We proceed by induction on the greatest prime divisor of $n$ (denoted by $\Omega_n$ in the following proof). We also denote the Chinese Remainder Theorem by CRT.
$\textbf{Initialisation:}$ For $n=1$, $a=0$ verifies $\gcd(a,n) = 1$, and $x=1$ works.
$\textbf{Induction:}$ Let $K>1$ be such that for all $n$ such that $\Omega_n<K$, for all $a\in \mathbb{N}^*$, $$\gcd(a,n)=1 \;\Rightarrow\; x^x\equiv a\pmod{n}\quad \text{has a solution.}$$
We take $n=p^k\cdot m$, where $\Omega_n = p$ is the smallest prime bigger than $K$, $k\geq 1$, and $\Omega_m < K$.
We also take $a\in \mathbb{N}$ which is relatively prime to $n$. We have $\gcd(a,m)=1$.
Moreover, let $r$ be the product of $q^{v_q(p-1)}$ where $q$ is a prime number that divides both $m$ and $p-1$.
$\gcd\left(\frac{p-1}{r},m\right) = 1$ so we can take $a\equiv 1\pmod{\frac{p-1}{r}}$. Applying CRT, we get that $\gcd(a,(p-1)m)=1$.
As $\Omega_{(p-1)m}<K$, there exists $x\in\mathbb{N}^*$ such that $x^x\equiv a\pmod{(p-1)m}$ $\color{red}{(i)}$.
We are now looking for $y\equiv x\pmod{(p-1)m\cdot\varphi((p-1)m)}$ $\color{red}{(ii)}$ such that $y^y\equiv a\pmod{p^k}$.
Notice that $\color{red}{(ii)}$ implies $y^y\equiv a\pmod{m}$.
$\color{red}{(ii)}$ also implies that $y\equiv x\pmod{p-1}$, so $y^y\equiv a\pmod{p}$ is equivalent to $y^x\equiv a\pmod{p}$. Considering that $\gcd(x,p) = 1$, $y\mapsto y^x$ is a bijection modulo $p$, so there exists a unique $y_1$ modulo $p$ such that $y_1^{y_1}\equiv a\pmod{p}$.
We are now looking for $y$ such that $y\equiv x\pmod{(p-1)m\cdot\varphi((p-1)m)}$ and $y\equiv y_1\pmod{p}$.
The two conditions $y\equiv x\pmod{p-1}$ and $y\equiv y_1\pmod{p}$ merge into $y\equiv a_1\pmod{p(p-1)}$ where $x_1$ is inversible modulo $p(p-1)$ and is given by CRT. $y^y\equiv a\pmod{p^2}$ becomes $y^{x_1}\equiv a\pmod{p^2}$, which has a unique solution $y_2$ (again, because $x_1$ is relatively prime to $p^2$, so $y\mapsto y^{x_1}$ is a bijection from and to the group of invertibles modulo $p^2$.)
Following this process, we find $y_1,y_2,\dots, y_k$, with at last $$
y\equiv y_k\pmod{p^k}\quad \Rightarrow\quad y^{y}\equiv a\pmod{p^k}$$
Using CRT (remember that we already have $y^y\equiv a\pmod{m}$), we are able to construct $y$ such that $y^y\equiv a\pmod{p^k\cdot m = n}$, which concludes the proof.
Thanks for reading and sorry for my bad english.
Best Answer
Instead of using the primitive root $2$, note that $7$ is a primitive root mod 11, because $7^2\equiv 5$ and $7^5\equiv -1$. So using 7-indices is a sensible move. We list the values: $$ \begin{array}{c|c|c} \hline x\pmod{11} & 5x^3\equiv 7^x\pmod{11} & x\pmod{10} \\\hline 1 & 5&2\\ 2 & 7&1\\ 3 & 3&4\\ 4 & 1&0\\ 5 & 9&8\\ 6 & 2&3\\ 7 &10&5\\ 8 & 8&9\\ 9 & 4&6\\ 10& 6&7\\\hline \end{array} $$ So combining the outer columns, you know ten solutions $x\pmod{110}$ and you know $x\equiv 2\pmod{3}$. Hence you can write down the ten solutions mod 330, and determine the solutions between 0 and 110.