This question is from Greuel et al., Introduction to Singularities and Deformations Exercise 2.1.3.
Consider the unfolding $$f_t(x,y,z)=x^p+y^q+z^r+txyz,\quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1.$$ Show that for all $t,t'\neq 0$ and $t\neq t'$, $f_t$ and $f_{t'}$ are contact-equivalent but not right-equivalent.
My attempt: I already proved that the two are contact-equivalent by explicitly finding an automorphism $\varphi$ of $\mathbb{C}\langle x,y,z\rangle$ such that there is a unit $u\in \mathbb{C}\langle x,y,z\rangle$ satisfying $\varphi(f_t)=uf_{t'}$. What I'm stuck is to prove that they are not right-equivalent. I once tried to prove that the Milnor number of $f_t$ and $f_{t'}$ are different, but they turned out to be equal, so I have to find another way. Does anyone have ideas?
Thanks in advance!
I don't have an account, so I can't post this as a comment. But it seems that there is a counter-example to the second part of this question. In particular, consider $$f_t(x, y, z) = x^6 + y^6 + z^6 + txyz$$ with $t = 1$ and $t' = -1$. If we define an automorphism $\varphi$ of $\mathbb{C}\langle x, y, z \rangle$ by \begin{align} x &\mapsto e^{i\pi/3}x, \\ y &\mapsto e^{i\pi/3}y, \\ z &\mapsto e^{i\pi/3}z, \end{align} then we have that \begin{align} \varphi(f_t) &= \varphi(x^6 + y^6 + z^6 + xyz) \\ &= (e^{i\pi/3}x)^6 + (e^{i\pi/3}y)^6 + (e^{i\pi/3}z)^6 + (e^{i\pi/3}x)(e^{i\pi/3}y)(e^{i\pi/3}z) \\ &= x^6 + y^6 + z^6 - xyz \\ &= f_{t'} \end{align} Thus, in this case $f_t \overset{r}{\sim} f_{t'}$.
As a further comment, it would not have been possible to use the Milnor number to distinguish the $f_t$ either. The book mentions that over a characteristic $0$ field such as $\mathbb{C}$, contact equivalence implies that the Milnor number is also equal. Since we already know that $f_t \overset{c}{\sim} f_{t'}$ for all $t$, $t' \neq 0$, we cannot use the Milnor number to show that $f_t \not\overset{r}{\sim} f_{t'}$.