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:

1. 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.

2. 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.

3. 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 $\mathrm{O}(8)$.