Let $\theta_a: S_3 \to S_3$ be the function given by $\theta_a(g) = aga^{-1}$. Let $a \in S_3$ and $\theta_a: S_3 \to S_3$ be the function given by $\theta_a(g) = aga^{-1}.$ Show that these define 6 different isomorphisms of $S_3.$
I know from the conjugation of permutation that $\sigma(1 2 3 ... n)\sigma^{-1} = (\sigma(1)\sigma(2)\sigma(3)...\sigma(n)).$ I also know  there are 6 elements in S3, which are e,(1 2),(1 3),(2 3),(1 2 3) and (1 3 2). (1 2 3)=(1 3)(1 2). But I really don't know how to prove this statement. 
 A: @Tsemo's answer may be suitable for you when you've studied a bit more group theory; my impression is that as of now, it's not something you're quite ready for, so I'll take a more basic approach. First, I'm going to use $a$ to denote $(1~2~3)$, so that I don't have to keep writing out permutations, and $b$ to denote $(1~2)$. I'll use $e$ for the identity permutation. It's not too hard to check that
$$
a^3 = b^2 = e
$$
and a little checking shows that 
$$
a b = ba^2
$$ (unless I've gotten it backwards and $ba = a^2b$, in which case I should have written $a = (3 ~ 2 ~ 1)$.)
Now let's look at $\theta_e, \theta_a, \theta_{a^2}$, and so on. I'll start out by showing that $\theta_e$ is different from $\theta_a$. How? I'll find an element $g$ with the property that 
$$
\theta_a(g) \ne \theta_e(g)
$$
for as you observed, two functions on $S_3$ are equal only if they take the same value on every element of $S_3$. 
I'm going to look at $g = b$, so let's look at 
$$
\theta_e(b) = e b e^{-1} = e b e = b.
$$
Now what's $\theta_a(b)?$ well, because $a^3 = e$ we see that $a \cdot a^2 = e$, so $a^{-1} = a^2$. So 
$$
\theta_a(b) = a b a^{-1} = a b a^2 = a (ab)
$$
where that last equality comes from the $ba^2  = ab$ that I wrote above. That further simplifies to $\theta_a(b) = a^2b$, which is different from $b$, so $\theta_e$ and $\theta_a$ are different. 
You can now do the same thing for each of the other $\theta_u$, for each element $u$ of the group $S_3$. You need to find something to show that each of these is different from all the others that you've encountered. Probably making a table of values will be your best bet. 
Best of luck, and when you're done, you'll see why Tsemo's compact answer is so very useful as a way of avoiding drudgery. :) 
A: Hint: Show that $\theta_{aa'}=\theta_a\circ\theta_{a'}$, it is a morphism of groups. Then remark that $\theta_a=Id$ implies that $a$ is an element of the center of $S_3$, since the center of $S_3$ is the neutral element, deduce that $\theta$ is injective.
