Q: simple question on conjugacy classes if $ \;G = \langle a,b\;|\; a^9 = b^3 = 1, bab^{-1} = a^4\rangle\; $ of order $\;27\;$
Then how would i show that $b$ is conjugate to $ba^3$
I have been fiddling around with this for ages and cannot come to a conclusion:
I showed that $a^3$ is its own conjugacy class as any power of $a$ and $a^{-1}$ on either side leaves $a^3$ unchanged and by multiplying both sides by $b$ you get $(a^3)^4$ = $a^{12}=a^3$.

A: Notice that $a^{-1}ba=a^3b$. Then,
$$(ba^{-1})b(ba^{-1})^{-1}=b(a^{-1}ba)b^{-1}=b(a^3b)b^{-1}=ba^3.$$
