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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 % $N(\phi(g))=\tau(\phi(g))\phi(g)=\tau(\Pi_{i=1}^n\phi(v_i))\phi(g)=\phi(\tau(g))\phi(g)=\phi(\tau(g)g)=\phi(N(g))=N(g)$. \begin{eqnarray*} N(\phi(g))&=&\tau(\phi(g))\phi(g)\\ & =&\tau(\phi(\Pi_{i=1}^nv_i))\phi(\Pi_{i=1}^nv_i)\\ & =&\tau(\Pi_{i=1}^n\phi(v_i))\phi(\Pi_{i=1}^nv_i)\\ & =&\Pi_{i=1}^n\phi(v_{n-i})\phi(\Pi_{i=1}^nv_i)\\ & =&\phi(\Pi_{i=1}^nv_{n-i})\phi(\Pi_{i=1}^nv_i)\\ & =&\phi(\Pi_{i=1}^nv_{n-i}\cdot\Pi_{i=1}^nv_i)\\ & =&\phi(\tau(g)g)\\ & =&\phi(N(g)) \end{eqnarray*} 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 $Pin(0,4)$ with norm $-1$.