Conjugate in Symmetry Group

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I have found the following question in one of my old assignments. I think it is wrong!
Question. Let $\psi = (i_1 i_2 \dots i_k)$ a $k$-cycle in $S_n$.Prove that: $$|N_{S_n}(\langle \psi \rangle):C_{S_n}(\psi)|=\varphi(n)$$ My problem is here! Shouldn't it be $\varphi(k)$ instead of $\varphi(n)$?

My attempt:
If $\beta \in N_{S_n}(\langle \psi \rangle)$ this means that: $$\beta \psi {\beta}^{-1} = {\psi}^m \;\text{and}\; (m,k)=1$$ Am I wrong here?
So know we can count the number of $N_{S_n}(\langle \psi \rangle)$. $$\beta \psi {\beta}^{-1}=(\beta(i_1) \beta(i_2) \dots \beta(i_k))=\psi^m$$ This means that with knowing only $\beta(i_1)=i_t$, then we have the value of $\beta(i)$ for all $i$. Thus, we can have only $k.(n-k)!$ elements that $\beta \psi {\beta}^{-1}=\psi^m$. We use the fact that the $\psi^m$ are $k-$cycles with different arrangement of $i_1, \dots, i_k$. As a result: $$|N_{S_n}(\langle \psi \rangle)|=\varphi(k).k.(n-k)!$$ We can conclude similarly as above that: $$|C_{S_n}( \psi )|=k.(n-k)!$$

Am I missing something? or is there a misspelling in the question?