It was in the test. I did: $Aut(\mathbb Z_{276}) =~ Aut(\mathbb Z_{23}) \times Aut(\mathbb Z_4) \times Aut(\mathbb Z_3) =~ U(\mathbb Z_{23}) \times U(\mathbb Z_3) \times U(\mathbb Z_4) =~ \mathbb Z_{11} \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ ,
Finding that $5$ is primitive root of $23$, so $U(\mathbb Z_{23})$ is cyclic of order $22$ and so isomorphic to $\mathbb Z_{22}$ and so to $\mathbb Z_{11} \times \mathbb Z_2$ because GCD$(11,2)=1$. The problem was to find a Sylow subgroup of order $2^3$ ($11\times2\times2\times2$ is the order of my group) but I didn't know why I couldn't find it. Help me please.
The answer depends on what you mean by "find".
As you have noticed, $$ Aut(\mathbb Z_{276}) \cong U(276) \cong U(23) \times U(3) \times U(4) \cong \mathbb Z_{11} \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 $$ Therefore, the $2$-Sylow subgroup of $Aut(\mathbb Z_{276})$ corresponds to $0 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$.
Thus, abstractly, the $2$-Sylow subgroup of $Aut(\mathbb Z_{276})$ is $C_2 \times C_2 \times C_2$.
If you want to describe this subgroup explicitly in $Aut(\mathbb Z_{276})$, you need to find the automorphisms of $\mathbb Z_{276}$ of order $2$. This can be done by solving $a^2 \equiv 1 \bmod 276$. Each solution gives an automorphism $x \mapsto ax$. You'll find that $a= \pm 1, \pm 47, \pm 91, \pm 137$.
Alternatively, you can always write down explicitly the three group isomorphisms above.