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$\sigma =\bigg(\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ 3 & 4 & 5 & 7 & 9 & 2 & 8 & 6 & 1 & 11 & 10 \end{matrix}\bigg) \in S_{11}$

Find $ \tau \in S_{11} $ for which: $\tau^{2011}=\sigma$

I do really have no idea on how to do that. Can you please help me?

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    To serve as a reference for the excellent answer of @Matthias Klupsch. The decomposition of $\sigma$ into product of disjoint cycles is: $$\sigma =\bigg(\begin{matrix} 1 & 3 & 5 & 9 \\ 3 & 5 & 9 & 1 \end{matrix}\bigg) \bigg(\begin{matrix} 2 & 4 & 7 & 8 & 6 \\ 4 & 7 & 8 & 6 & 2 \end{matrix}\bigg)\bigg(\begin{matrix} 10 & 11 \\ 11 & 10 \end{matrix}\bigg)$$ thus $\sigma$ has order LCM$(4,5,2)=20$. – 2017-01-23

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As we already have seen in other answers, $\sigma$ has order $20$. Now $1 = 11 \cdot 2011 - 1106 \cdot 20 $ and so we have $$ \sigma = \sigma^{11 \cdot 2011 - 1106 \cdot 20} = (\sigma^{11})^{2011} \cdot (\sigma^{20})^{-1106} = (\sigma^{11})^{2011}$$ Thus $\tau = \sigma^{11}$ does the job.