Let Sch be the category of schemes, and Ring be the category of commutative rings, then Spec and the global sections functor $\Gamma$ are adjoint between these two categories. Ie, for any scheme $X$ and ring $A$ there is a natural bijection $$Hom_{Sch}(X,{\rm Spec}\,A) = Hom(A,\Gamma(X))$$ If $A = \Gamma(X)$, then we get $$Hom_{Sch}(X,{\rm Spec}\,\Gamma(X)) = Hom(\Gamma(X),\Gamma(X))$$ Thus, the identity on the right side corresponds to some special morphism $X\rightarrow{\rm Spec}\,\Gamma(X)$.
Now, the definition of this morphism is given in rather grueling detail in http://stacks.math.columbia.edu/tag/01HX
My question is - how should I think about this intuitively? I suppose one obstacle to thinking about this is the lack of examples. The only examples I'm comfortable with are either affine or projective curves, and in either case this morphism is trivial.
The map $z:Hom_{Sch}(X,{\rm Spec}\,\Gamma(X))$ corresponding to the identity $Id:Hom(\Gamma(X),\Gamma(X))$ is easy to interpret geometrically: it associates to a point $x\in X$ the set of global regular functions on $X$ which are zero at $x$.
That set is a prime ideal $j_x\subset\Gamma(X)$, hence corresponds to a point of the target affine scheme ${\rm Spec}\,\Gamma(X)$.
(Note carefully: saying that $f\in \Gamma(X)$ is zero at $x$ means that the class of its germ at $x$ is zero : $[f_x]=0\in \mathcal O_{X,x}/\mathfrak m_x=\kappa(x)$.)
The beauty of scheme theory is that the above description includes the case where $x$ is not closed: then you get the prime ideal $j_x$ corresponding to the set of regular functions zero on a more classical-looking variety, namely the closure $\overline {\{x\}}$ of the point $x$.
In some cases this map $z$ is quite uninteresting: if $X$ is an integral projective (or even complete) scheme, then all $j_x=(0)$ so that the map $z$ sends all points of $X$ to the generic point of ${\rm Spec}\,\Gamma(X)$, in other words it is useless!
The opposite extreme is the case when $X$ is affine: then the map $z$ is an isomorphism of schemes.
The "intermediate" cases can be quite instructive: for example if $X=\mathbb A^2_k\setminus \{\langle x,y\rangle\}\subset \mathbb A^2_k=\operatorname {Spec} k[x,y]$, the punctured affine plane over the field $k$, the map $z$ is almost surjective: its image is all of $\mathbb A^2_k$ except the deleted point $\langle x,y \rangle$.
This constitutes a proof that $X$ is not affine !