Looking at Sylow questions on $GL_2(\mathbb F_3)$. we have that $Q$ is the unique $2$-Sylow of $N=SL_2(\mathbb F_3)$. $|Q|=8=2^3$ hence by the classification of groups of order $p^3$, we have 5 possibilies for $Q$: $\;C_8,\;C_4\times C_2,\;C_2^3,\;D_4,\;Q_8$. But looking at On $GL_2(\mathbb F_3)$, we see that we have only one element of order $2$ in $N$, hence the same holds for $Q$, then we must exclude $C_4\times C_2,\;C_2^3,\;D_4$.
Hence $Q=C_8$ or $Q=Q_8$. In order to exclude the case $Q=C_8$ my teacher said that even though $Q\unlhd N$, "$Q$ is not centralized by any $3$-Sylow subgroup of $N$" (there are four $3$-Sylow subgroups of $N$, see again Sylow questions on $GL_2(\mathbb F_3)$.), and so $Q$ has an automorphism of order $3$.
I think that by "$Q$ is not centralized by any $3$-Sylow subgroup of $N$", my teacher mean that although $Q\unlhd N$, and thus $Q^g=Q\;\;\forall g\in N$, calling $B$ a $3$-Sylow of $N$, it's not true that $qb=bq\;\;\forall q\in Q,\;\;\forall b\in B$ (first question: how can I see this?), hence we can define a $\psi\in Aut(Q)$ defined by $\psi(q)=q^b$; and the second question is: how can I prove that $\psi$ has order $3$ (the order of an automorphism is defined as the minmum $n\in\mathbb N$ s.t. $\psi^n=id_Q$)? And if $\psi$ is not the right automorphism, (third question) how can I define an automorphism of $Q$ of order $3$?
Proving that there exists an automorphism of order $3$, we can argue as follows: if by contradiction $Q=C_8$, then $Aut(C_8)\simeq U(\mathbb Z_8)$ hence $|Aut(C_8)|=4$, thus it can't contain any automorphism of order $3$. Then I can conclude that $Q=Q_8$.
Thank you all