Let $S^d$ denote the $d$-sphere. The only non-trivial cohomology groups are $H^0(S^d;\mathbb Z_2)= \mathbb Z_2$ generated by $1$ and $H^d(S^d;\mathbb Z_2)= \mathbb Z_2$ generated by the fundamental class $u$. I want to write all the Steendod squares on the sphere, that is all the group homomorphisms $Sq^i:H^n(S^d;\mathbb Z_2)\to H^{n+i}(S^d;\mathbb Z_2)$ satisfying some axioms.
I claim that there are only three of them:
$Sq^0:H^0(S^d;\mathbb Z_2)\to H^0(S^d;\mathbb Z_2)$ which is the identity by the first axiom;
$Sq^d:H^0(S^d;\mathbb Z_2)\to H^d(S^d;\mathbb Z_2)$ which sends $1$ to $u$;
$Sq^0:H^d(S^d;\mathbb Z_2)\to H^d(S^d;\mathbb Z_2)$ which is the identity.
Is this correct?