Suppose that $S$ is a subset of the group $G$. I would like to show that $g^{-1} \langle S \rangle g = \langle g^{-1} S g \rangle$. In my book, $\langle S \rangle := \bigcap_{S \subseteq H \\ H \le G} H$. Since conjugation is automorphism, it is a fortiori an injection. Hence,
$$g^{-1} \langle S \rangle g = g^{-1} \bigcap_{S \subseteq H \\ H \le G} H g$$
$$=\bigcap_{S \subseteq H \\ H \le G} g^{-1} Hg$$
Here is where things begin to feel uncertain. I want to say that, because $S \subseteq H$ iff $g^{-1}Sg \subseteq g^{-1}Hg$, then I can just write
$$\bigcap_{S \subseteq H \\ H \le G} g^{-1} Hg = \bigcap_{g^{-1}Sg \subseteq g^{-1}Hg \\ H \le G} g^{-1} Hg = \langle g^{-1}Sg \rangle$$
But this doesn't feel quite right to me...Perhaps someone could clarify why this is or is not right.
The easiest way to prove $g^{-1} \langle S \rangle g = \langle g^{-1} S g \rangle$ is to use the characterization of $ \langle X \rangle$ as the set of all words in $X \cup X^{-1}$.
Then every $a \in g^{-1} \langle S \rangle g$ can be written $$ a = g^{-1} x_1 x_2 \cdots x_n g = g^{-1} x_1 g g^{-1} x_2 \cdots g^{-1} x_n g \in \langle g^{-1} S g \rangle $$ Conversely, every $b \in \langle g^{-1} S g \rangle$ can be written $$ b = y_1 y_2 \dots y_n = (g^{-1} x_1 g) (g^{-1} x_2 g) \cdots (g^{-1} x_n g) = g^{-1} x_1 x_2 \cdots x_n g \in g^{-1} \langle S \rangle g $$