Let G be a group, Let $I$ is indexing set. $H_i$ be a subgroup of $G$. Prove that for any $g \in G$,
$$ g (\cap_{i \in I} H_i )g^{-1} = \cap_{i \in I} g H_i g^{-1} $$
some set notations
$$ \begin{aligned} g \cap _{i \in I} H_i g^{-1} &=\{ g hg^{-1} : h \in \cap_{i \in I} H_i \} \\ gH_ig^{-1} &= \{ g hg^{-1}: h \in H_i \} \end{aligned} $$
$\Leftarrow ]$ $( \cap_{i \in I} g H_i g^{-1}) \subseteq g( \cap_{i \in I H_i})g^{-1}$
Assume $x \in \cap_{i \in I } gH_ig^{-1}$
Now
$$\begin{aligned} x \in gH_1g^{-1} \cap \dots \cap gH_i g^{-1} \\x \in gH_1g^{-1} \wedge \dots \wedge x \in gH_i g^{-1} \end{aligned} $$
so $$\begin{aligned} x &=g h_1 g^{-1} && h_1 \in H_1 \\ &= g h_2 g^{-1 } && h_2 \in H_2 \\ &\vdots \\ &=g h_i g^{-1} && h_i \in H_i \end{aligned} $$
Missing argument, theorem, prop where means h is all H's
Therefore $$\begin{aligned} x &\in \{ ghg^{-1} : h\in \cap_{i \in I} h_i\} \\ x &\in g (\cap_{i \in I} H_i)g^{-1} \end{aligned} $$
$\Rightarrow]$ $ \left( g(\cap_{i \in I} H_i)g^{-1} \subseteq g H_i g^{-1} ) \right )$
$x\in g (\cap_{i \in I} H_i)g^{-1}$
making of the form
$$ x = ghg^{-1} \text{ where } h \in \cap_{i \in I} H_i$$
Now, $$\begin{aligned} x &= ghg^{-1} \text{ where } h \in H_1 \\ &\vdots \\ x &= ghg^{-1} \text{ where } h \in H_i \end{aligned} $$
so $$\begin{aligned} x &\in gH_1g^{-1} \\ &\vdots \\ x &\in gH_ig^{-1} \end{aligned} $$
that is $$\begin{aligned} x &\in gH_1 g^{-1 } \wedge \dots \wedge x \in gH_i g^{-1 } \\x &\in gH_1 g^{-1 } \cap \dots \cap gH_i g^{-1 } \\ x &\in \cap_{i \in I} gH_i g^{-1} \end{aligned} $$
case for when the $H_i$ are normal
Proposition 13 sec 4.4 Automorphism Dummit's
Let $H$ be a normal subgroup of group $G$. Then $G$ acts by conjugation on $H$ as automorphism of $G$
call $$H = \cap_{i \in I} H_i $$
$$\begin{aligned} x &\in g ( \cap _{i \in I}) g^{-1} \\ x &\in g (H_1 \cap H_2 \cap \dots \cap H_i)g^{-1} \\ x &\in g Hg^{-1} \end{aligned} $$
From the proposition, This conjugation is an automorphism so $x \in g H_i g^{-1} \Leftrightarrow x \in H_i$. Now $x\in H \Leftrightarrow x \in gHg^{-1}$
so $x\in H$ means that
$$\begin{aligned} x &\in H_1 \cap \dots \cap H_i \\ x &\in H_1 \wedge \dots \wedge x \in H_i \end{aligned} $$
again automorphism means that $\forall g\in G$
$$\begin{aligned} x &\in g H_1 g^{-1} \wedge \dots \wedge x\in g H_ig^{-1} \\ x&\in g H_1 g^{-1} \cap \dots \cap g H_ig^{-1} \\ x&\in \cap_{i \in I} gHg^{-1} \end{aligned} $$
\textbf{Case when the H's are not normal}
unless i have missed something you may be over-thinking this one. something like the following sequence of equivalences should be all that is required
$$ \begin{align} x \in g(\cap_i H_i)g^{-1} &\iff g^{-1}xg \in \cap_i H_i \\&\iff \forall i.g^{-1}xg \in H_i \\&\iff \forall i.x \in gH_ig^{-1} \\&\iff x \in \cap_i gH_ig^{-1} \end{align} $$