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.