If $f$ is an automorphism of $S_3$, prove that there exists $\sigma\in S_3$ such that $f(\tau)=\sigma\tau\sigma^{-1}$ for every $\tau\in S_3$.
I am not sure what the question is asking. I know that a 3-cycle can be generated from two 2-cycles. So we only need to consider the three transpositions. So am I suppose to consider what $f(\tau)$ equals to by plugging into $\tau=$ $(12), (13), (23)$, and for each $f(\tau)$, I am suppose to look for a specific $\sigma$ and then go through it for each $\tau$.
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I know that $(13)(12)(31)=(23)$ and $(12)(13)(12)=(23)$. I also done the following calculations:
$(23)(12)(23)=(13)$
$(13)(12)(31)=(23)$
$(12)(12)(12)=(12)$
$(123)(12)(132)=(23)$
$(132)(12)(123)=(13)$
$(13)(23)(31)=(12)$
$(12)(23)(21)=(13)$
$(23)(23)(23)=(23)$
$(123)(23)(132)=(13)$
$(132)(23)(123)=(12)$
$(12)(13)(12)=(23)$
$(23)(13)(22)=(12)$
$(13)(13)(13)=(13)$
$(123)(13)(132)=(12)$
$(132)(12)(123)=(23)$
$(132)(123)(123)=(123)$
$(123)(123)(132)=(123)$
$(12)(123)(21)=(132)$
$(23)(123)(32)=(132)$
$(13)(123)(31)=(132)$
$(132)(132)(123)=(132)$
$(123)(132)(132)=(132)$
$(12)(132)(21)=(123)$
$(13)(132)(31)=(123)$
$(23)(132)(32)=(123)$
Can someone clarify the question for me please. Thank you in advance
Edit: Since the mods chose to close the question, I would like to give an example to see if this is what the question is asking. Is the question asking for me to do the following: for $\tau=(12)$, we have $f(12)=$ either $(12), (13)$ or $(23)$. If it is $\tau=(12)$, then $\sigma=(12)$ if $f(12)=(13)$, then $\sigma=(23)$ and if $f(12)=(23)$, then $\sigma=(13)$. If the example is not sufficient, what else am I asked to do in the problem. I am really confused over the wording of the question.