$|G|=11$ operates on $\mathbb{Z} / 5\mathbb{Z}\times \mathbb{Z} / 5\mathbb{Z}$ implies at least one point fixpoint Can someone please help me with the following: A group of order $11$ operates on $\mathbb{Z} / 5\mathbb{Z}\times \mathbb{Z} / 5\mathbb{Z}$. I have to show that it has at least one fix point. Can someone help me please with it or give me at least some hint ?
I have an exam tomorrow so please give me some hint!

Robert's answer already wrapped up this question, but you may be interested in proving the following general proposition, which solves at once your question:
Proposition: If $\,G\,$ is a finite $\,p-$group and $\,X\,$ is a non-empty finite set upon which $\,G\,$ acts, then
$$|X|=\left|X^G\right|\pmod p\;\;,\;\;\text{with}\;\;X^G:=\{x\in X\;gx=x\,\,\forall g\in G\}=\{x\in X\;;\;|\mathcal Orb(x)|=1\}$$
