You can work out the answers to these questions using the material in Chapter 11 of the book Spinors and Calibrations by F. Reese Harvey. You will also need to recall that, for $N\not=4$, the group of automorphisms of $\mathrm{Spin}(2N)$ is $\mathrm{O}(2N)$ rather than $\mathrm{SO}(2N)$. The answers depend to some extent on $N$ modulo $4$.
The point is that $\mathrm{Spin}(2N)$ is represented in $\mathbb{C}^{2^{N-1}}$ as either the full spinor space (when $N$ is odd) or a semi-spinor space (when $N$ is even), and the commuting ring of this representation (which may include complex conjugation) depends on $N$ modulo $4$.
For example, when $N=4$, the representation of $\mathrm{Spin}(8)$ is actually the complexification of a 8-dimensional real representation of $\mathrm{Spin}(8)$, whose image in $\mathrm{O}(8)$ is just a double cover onto $\mathrm{SO}(8)$. (In particular, this semi-spinor representation is not faithful.) Thus, for clarity, let me call the image of this representation $\mathrm{SO}'(8)\subset\mathrm{O}(8)\subset\mathrm{U}(8)$. The normalizer of $\mathrm{SO}'(8)$ in $\mathrm{O}(8)$ is $\mathrm{O}(8)$, and conjugation by elements of $\mathrm{O}(8)$ induce all of the automorphisms of $\mathrm{SO}'(8)\simeq \mathrm{SO}(8)$. Noting that the centralizer of $\mathrm{SO}'(8)$ in $\mathrm{U}(8)$ is simply $S^1 = \{\lambda I_8 \ |\ |\lambda|^2=1\ \}$, it now follows easily that the normalizer of $\mathrm{SO}'(8)$ in $\mathrm{U}(8)$ is simply $S^1{\cdot}\mathrm{O}(8)$, which has two components.
Meanwhile, when $N=5$, the group $\mathrm{Spin}(10)$ embeds into $\mathrm{SU}(16)$ and this irreducible representation on $\mathbb{C}^{16}$ is not the complexification of a 16-dimensional real representation. Moreover, the conjugate representation on $\mathbb{C}^{16}$ is not isomorphic (as a complex representation) to the given representation. In particular, an element of the normalizer of $\mathrm{Spin}(10)$ in $\mathrm{SU}(16)$ can only induce an inner automorphism of $\mathrm{Spin}(10)$. Meanwhile, the centralizer of $\mathrm{Spin}(10)$ in $\mathrm{U}(16)$ is $S^1$, the multiples of the identity. Thus, the normalizer of $\mathrm{Spin}(10)$ in $\mathrm{U}(16)$ is $S^1{\cdot}\mathrm{Spin}(10)$, which is connected.
In general, here is some information about the spin representations that can be gleaned from Harvey's book:
For $n>0$, the spin representation $\rho:\mathrm{Spin}(4n{+}2)\to\mathrm{SU}(2^{2n})$ is faithful and irreducible (even as a real representation). Moreover, the center of $\mathrm{Spin}(4n{+}2)$, which is isomorphic to $\mathbb{Z}_4$, is mapped under $\rho$ to $\{\,\lambda I_{2^{2n}}\ |\ \lambda^4 = 1\,\}$, which lies in the center of $\mathrm{SU}(2^{2n})$. In particular, if $N\subset \mathrm{SU}(2^{2n})$ is the normalizer of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$ in $\mathrm{SU}(2^{2n})$ then conjugation by an element $g\in N$ is the identity on the center of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$, so it represents an inner automorphism of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$ and hence can be written in the form $g = \rho(h)z$, for some $h\in\mathrm{Spin}(4n{+}2)$ and some $z\in\mathrm{SU}(2^{2n})$ that lies in the centralizer of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$. Because the representation $\rho$ is irreducible, this centralizer must be a multiple of the identity, i.e., $z = \lambda I_{2^{2n}}$ where $\lambda^{2^{2n}} = 1$. Thus, $N$ is the product in $\mathrm{SU}(2^{2n})$ of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$ with the center of $\mathrm{SU}(2^{2n})$. In particular, it follows that the normalizer of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$ in $\mathrm{U}(2^{2n})$ is the product of $\rho\bigl(\mathrm{Spin}(4n{+}2)\bigr)$ with the center of $\mathrm{U}(2^{2n})$, a group isomorphic to $S^1$. In particular, the normalizer in the full unitary group is connected.
For $n>0$, the spin representation $\rho = (\rho_+,\rho_-):\mathrm{Spin}(8n{+}4)\to\mathrm{Sp}(2^{4n})\times\mathrm{Sp}(2^{4n})$
is faithful, but each of the semi-spin representations $\rho_\pm:\mathrm{Spin}(8n{+}4)\to\mathrm{Sp}(2^{4n})$, while irreducible (even as a real representation), is a double cover onto its image in $\mathrm{Sp}(2^{4n})$. In fact, the center of $\mathrm{Spin}(8n{+}4)$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ in such a way that the kernel of $\mathrm{Spin}(8n{+}4)\to \mathrm{SO}(8n{+}4)$ is $\{(\pm 1,\pm1)\}\subset\mathbb{Z}_2\times\mathbb{Z}_2$, while the kernel of $\rho_+$ is $\{(\pm 1,1)\}\subset\mathbb{Z}_2\times\mathbb{Z}_2$ and the kernel of $\rho_-$ is $\{(1,\pm1)\}\subset\mathbb{Z}_2\times\mathbb{Z}_2$. I don't know a universally agreed-on notation, but some writers use $\mathrm{SO}'(8n{+}4)\subset\mathrm{Sp}(2^{4n})$ for $\rho_+\bigl(\mathrm{Spin}(8n{+}4)\bigr)$ and $\mathrm{SO}''(8n{+}4)\subset\mathrm{Sp}(2^{4n})$ for $\rho_-\bigl(\mathrm{Spin}(8n{+}4)\bigr)$. It is important to note that neither $\mathrm{SO}'(8n{+}4)$ nor $\mathrm{SO}''(8n{+}4)$ have outer automorphisms. Consequently, the normalizer of $\mathrm{SO}'(8n{+}4)$ in $\mathrm{Sp}(2^{4n})$ consists of the product of $\mathrm{SO}'(8n{+}4)$ with its centralizer in $\mathrm{Sp}(2^{4n})$. But, since $\rho_+$ is irreducible as a real representation, its centralizer in $\mathrm{SO}(2^{4n+2})$ is $\mathrm{Sp}(1)$, of which, only its center lies in $\mathrm{Sp}(2^{4n})$ and hence in $\mathrm{SO}'(8n{+}4)$. Thus, the normalizer of $\mathrm{SO}'(8n{+}4)$ in $\mathrm{Sp}(2^{4n})$ is just $\mathrm{SO}'(8n{+}4)$ itself. From this information, it is now easy to determine the normalizers of $\mathrm{SO}'(8n{+}4)$ in the larger groups $\mathrm{U}(2^{4n+1})$ and $\mathrm{O}(2^{4n+2})$. The story for $\mathrm{SO}''(8n{+}4)$ is similar.
For $n>0$, the spin representation $\rho = (\rho_+,\rho_-):\mathrm{Spin}(8n)\to\mathrm{SO}(2^{4n-1})\times\mathrm{SO}(2^{4n-1})$
is faithful, but each of the semi-spin representations $\rho_\pm:\mathrm{Spin}(8n)\to\mathrm{SO}(2^{4n-1})$, while irreducible with commuting ring $\mathbb{R}$, is a double cover onto its image in $\mathrm{SO}(2^{4n-1})$. In fact, the center of $\mathrm{Spin}(8n)$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ in such a way that the kernel of $\mathrm{Spin}(8n)\to \mathrm{SO}(8n)$ is $\{(\pm 1,\pm1)\}\subset\mathbb{Z}_2\times\mathbb{Z}_2$, while the kernel of $\rho_+$ is $\{(\pm 1,1)\}\subset\mathbb{Z}_2\times\mathbb{Z}_2$ and the kernel of $\rho_-$ is $\{(1,\pm1)\}\subset\mathbb{Z}_2\times\mathbb{Z}_2$. Again, I don't know a universally agreed-on notation, but some writers use $\mathrm{SO}'(8n)\subset\mathrm{SO}(2^{4n-1})$ for $\rho_+\bigl(\mathrm{Spin}(8n)\bigr)$ and $\mathrm{SO}''(8n)\subset\mathrm{SO}(2^{4n-1})$ for $\rho_-\bigl(\mathrm{Spin}(8n)\bigr)$. When $n>1$, neither $\mathrm{SO}'(8n)$ nor $\mathrm{SO}''(8n)$ have outer automorphisms. Consequently, the normalizer of $\mathrm{SO}'(8n)$ in $\mathrm{O}(2^{4n-1})$ consists of the product of $\mathrm{SO}'(8n)$ with its centralizer in $\mathrm{O}(2^{4n-1})$. But, since $\rho_+$ is irreducible with commmuting ring $\mathbb{R}$, its centralizer in $\mathrm{O}(2^{4n-1})$ is $\pm 1$ times the identity, which already lies in $\mathrm{SO}'(8n)$. Thus, the normalizer of $\mathrm{SO}'(8n)$ in $\mathrm{O}(2^{4n-1})$ is just $\mathrm{SO}'(8n)$ itself. The story for $\mathrm{SO}''(8n{+}4)$ is similar. Finally, when $n=1$, it turns out that $\mathrm{SO}'(8)\simeq \mathrm{SO}''(8)\simeq \mathrm{SO}(8)$ (because of triality), so these groups do have outer automorphisms, and so the normalizer of these groups in $\mathrm{O}(8)$ is $\mathrm{O}(8)$.
