Your initial claim about $Pin(4,0)\not\simeq
Pin(0,4)$ seems to be correct. In fact let $\phi:Pin(4,0)\rightarrow
Pin(0,4)$ be an (abstract) group isomorphism. 
When $p+q$ is even, 
every element $g$ of the Clifford-Lipschitz group $\Gamma(p,q)$ is a
product of some anisotropic elements $v_i, (i=1,\cdots,n)$ of $V$ (this follows from the
Cartan-Dieudonn\'e theorem).
Now let $g=v_1\cdots v_n\in Pin(4,0)$. 
By scaling we can still have $g=v_1\cdots v_n$ with  $N(v_i)=\pm1$,
hence we may assume that $v_i\in Pin(4,0)$.
So $\phi(g)=\Pi_{i=1}^n\phi(v_i)$.
We then obtain
$N(\phi(g))=\tau(\phi(g))\phi(g)=\tau(\Pi_{i=1}^n\phi(v_i))\phi(g)=\phi(\tau(g))\phi(g)=\phi(\tau(g)g)=\phi(N(g))=N(g)$.
But this is contradiction because considering the fact that $(4,0)$ is
positive definite, $N$ can take only positive values on $\Gamma(4,0)$.
But we have many elements of $\Gamma(0,4)$ with norm $-1$.