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I am trying to calculate $\text{Aut} (D_3)$, the automorphism group of the group of symmetries of the triangle. But I got stuck and now I have two questions about this.

Let me share my thoughts first:

First I noted that an isomorphism maps elements to elements of same order. Therefore, any $\varphi : D_3 \to D_3$ maps reflections to reflections and rotations to rotations. As candidates I got:

(1) the identity map $\varphi_1 (x) = x$

(2) the map $\varphi_2 (x) = x$ except $R_{120}\mapsto R_{240}$ and $R_{240}\mapsto R_{120}$

(3) the map $\varphi_3 (x) = x$ except $F_1 \mapsto F_2$ and $F_2 \mapsto F_1$

(4) the map $\varphi_4 (x) = x$ except $F_2 \mapsto F_3$ and $F_3 \mapsto F_2$

(5) the map $\varphi_5 (x) = x$ except $F_1 \mapsto F_3$ and $F_3 \mapsto F_1$

Then since $F_1 F_2 = R_{120}$, $\varphi_3 (R_{120}) = R_{240}$ and $\varphi_3 (R_{240}) = R_{120}$. Now I wanted to show that also $\varphi_2 (F_1) = F_2 $ and $\varphi_2(F_2) = F_1$ but unfortunately this doesn't have to be true since $\varphi_2$ could map $F_1$ to $F_3$ for example.

Now my question is:

Can one say something like we relabel the reflections and therefore $\varphi_2 (F_1) = F_2$?

If yes then all these are equal and there are only two automorphisms --

That I have found anyway because my next question is:

How can I be sure that I found all automorphisms of a given group? Are there upper and lower bounds on the number?

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    $\begingroup$ If you think you've identified all of the automorphisms, then you can try to prove that any automorphism must be equal to one of the automorphisms you've identified. This will just be a hard problem in general but in this case you can do some case analysis and it shouldn't be so bad. $\endgroup$ – Qiaochu Yuan Jan 17 '16 at 0:30
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    $\begingroup$ @QiaochuYuan Is there any way to know how many there are? $\endgroup$ – a student Jan 19 '16 at 0:07
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    $\begingroup$ You prove that you've found them all, one way or another. Again, in general this is hard. For example, it turns out that all of the automorphisms of the symmetric group $S_n$ are inner, except when $n = 6$. $\endgroup$ – Qiaochu Yuan Jan 19 '16 at 0:08
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    $\begingroup$ @QiaochuYuan Right, I meant to calculate their number before finding them. I understand that you're saying that's not possible (except in certain known cases). Thank you for your comments! $\endgroup$ – a student Jan 19 '16 at 1:49
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Let's characterize $D_3$ as the group generated by $r,t$ with $$r^3 = t^2 = (rt)^2 = e$$ $\hspace{10pt}$Suppose that $\phi:D_3 \to D_3$ is an automorphism.

Then, $\phi(t)$ must be of order two $\Rightarrow$ $\phi(t) \in \{t,rt, r^2 t\}$.

Similarly we can see that $\phi(r)$ must be of order three $\Rightarrow$ $\phi(r) \in \{r,r^2\}$.

Since these two choices determine $\phi$ (and any two are compatible), we have $2 \times 3 = 6$ possible choices for $\phi$.

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