Show $\epsilon e^{{i}{\vec\phi\over2}\cdot \vec{\sigma}^*} (-\epsilon) =e^{-{\vec\phi\over2}\cdot \vec\sigma} $ for Pauli matrices

Question

In pg. 76 of the Physics from Symmetry book, it was stated that the following relation is true:

$$\epsilon e^{{i}{\vec\phi\over2}\cdot \vec{\sigma}^*} (-\epsilon) =e^{-i{\vec\phi\over2}\cdot \vec\sigma} $$

where $\epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$, $\vec\phi$ is a constant vector and $\vec\sigma = (\sigma_x,\sigma_y,\sigma_z)$ represent the Pauli matrices.

But for a function $f(\epsilon \sigma_i \epsilon^{-1})$

It was stated that this was shown true by using the fact that $$\epsilon\sigma_i^*(-\epsilon)=-\sigma_i.$$

I had been able to prove that the first relation is true up to a second order expansion in Pauli matrices explicitly. However, is there an easier way to see that it is true generally without having to explicitly expand to the $n^\text{th}$ order?

Edit: I had tried the methods the comments and answer had suggested but is still unsuccessful. The main challenge I face is that in the expansion of $\epsilon e^{{i}{\vec\phi\over2}\cdot \vec{\sigma}^*} (-\epsilon)$ there will be cross terms like $\epsilon {\sigma_x^*}^n {\sigma_y^*}^m {\sigma_z^*}^l (-\epsilon)$ where $n,m,l$ are integers. If it can be shown that $\epsilon {\sigma_x^*}^n {\sigma_y^*}^m {\sigma_z^*}^l (-\epsilon) = (-1)^{n+m+l} {\sigma_x}^n {\sigma_y}^m {\sigma_z}^l $ , then the first relation will be true. However, I don't know how to show it.

Answer 1

$$\epsilon = i\sigma_2 = -\epsilon^{-1}, \\ \leadsto \epsilon \sigma_i \epsilon^ {-1}= -\sigma_i^*, $$ the conjugate representation.

As a similarity transformation it leads to $$ \epsilon \sigma_i^n \epsilon^ {-1}= (-\sigma_i^*)^n, \leadsto \\ f(\epsilon \sigma_i \epsilon^ {-1})= f(-\sigma_i^*), $$ for any function f regular at the origin, including the exponential.