Isomorphism of orthogonal groups 
1.5 Lemma.
  If $(V,b) \cong (V', b')$, then $\operatorname{O}(V,b) \cong \operatorname{O}(V',b')$.
Proof:
  If $\sigma \colon V \to V'$ is a bijective isometry, then $\tau \mapsto \sigma \tau \sigma^{-1}$ is an isomorphism of the orthogonal groups.
(Original image here.)

Since vector space are isomorphic there exist bijective isometry between those two vector space. I am confused whether $\tau$ is map from $V$ to $V$ or something else and how it become isomorphic.
 A: The text provides the explicit map of the isomorphism between the orthogonal groups namely
$$\tau \mapsto \sigma \tau \sigma^{-1}$$
There are 3 things you'd need to show to flesh out the details:


*

*If $\tau$ in the orthogonal group of $V$ then   $\sigma \tau \sigma^{-1}$ is in the orthogonal group $V'$ 

*$\tau \mapsto \sigma \tau \sigma^{-1}$ is a homomorphism

*$\tau \mapsto \sigma \tau \sigma^{-1}$ is bijective
You seemed most concerned about 3 in the comments so lets start with that. 

To SHOW 1)

$\tau \mapsto \sigma \tau \sigma^{-1}$ is injective
Suppose that $\sigma \tau \sigma^{-1}=I$ where $I$ is the identity transformation of $V'$ then $\tau=\sigma^{-1}I\sigma$. Pick an arbitrary element $v \in V$ then 
$$\begin{split}
\tau(v)&=(\sigma^{-1}I\sigma)(v)\\
&=\sigma^{-1}I(\sigma(v))\\
&=\sigma^{-1}(\sigma(v))\\
&=v\\
\end{split}$$
So $\tau$ is the identity transformation on $V$. Therefore the matp is injective.

$\tau \mapsto \sigma \tau \sigma^{-1}$ is surjective
Let $\alpha:V'\to V'$ be an invertible orthogonal transformation. Then consider $\tau:V \to V$ defined by $\tau=\sigma^{-1}\alpha\sigma$. It's clear that $\tau$ is invertible but we need to show it preserves the bilinear form $b$. 
Note that $\sigma:V \to V'$ being a linear isometry means that for all $a,b \in V$
$$
b(a,b)=b'(\sigma(a),\sigma(b))
$$
and for all $x,y \in V'$
$$
b'(x,y)=b(\sigma^{-1}(a),\sigma^{-1}(b))
$$
Also note since $\alpha$ is in the orthogonal group of $V'$
$$
b'(\alpha (x),\alpha(y))=b'(x,y)
$$
So let $v,w \in V$ and consider 
$$\begin{split}
b(\tau(v),\tau(w))&=b(\sigma^{-1}\alpha\sigma(v),\sigma^{-1}\alpha\sigma(w)) \\
&=b(\sigma^{-1}(\alpha\sigma(v)),\sigma^{-1}(\alpha\sigma(w))) \\
&=b'(\alpha\sigma(v),\alpha\sigma(w))\\
&=b'(\sigma(v),\sigma(w))\\
&=b(v,w)\\
\end{split}
$$
So $\tau$ preserves the bilinear form $b$. Note that usually $\sigma$ is assumed to be linear. Often it is called a linear isometry. I don't think this is true or even 1) is true if $\sigma$ is not linear. 

To SHOW 2)
$\alpha,\beta$ in the orthogonal group of $V$. Then 
$$
\sigma \alpha \beta \sigma^{-1}=(\sigma \alpha\sigma^{-1}) (\sigma\beta \sigma^{-1} )
$$ 


To SHOW 1)
Let $v,w \in V'$, tau in the orthogonal group of $V$
$$\begin{split}
b'(\sigma\tau\sigma^{-1}(v),\sigma\tau\sigma^{-1}(w)) 
&=b'(\sigma(\tau\sigma^{-1}(v)),\sigma(\tau\sigma^{-1}(w))) \\
&=b(\tau\sigma^{-1}(v),\tau\sigma^{-1}(w)) \\
&=b(\sigma^{-1}(v),\sigma^{-1}(w)) \\
&=b'(v,w)\\
\end{split}
$$
The linearity and invertibility of $\sigma\tau\sigma^{-1}$ follow from $\sigma,\tau,\sigma^{-1}$ each being linear and invertible.
