When does $\mathrm{End}(X)\cong X$? I noticed in lecture that for any field $\mathbb F$, take the one-dimensional vector space $V$ over $\mathbb F$ and write $\mathbb F\cong V\cong V^\ast=\mathrm{Hom}(V,\mathbb F)\cong\mathrm{Hom}(\mathbb F,\mathbb F)=\mathrm{End}(\mathbb F)$. This is a very basic case, so I'm just wondering, more generally, for what kinds of spaces $X$ is $\mathrm{End}(X)\cong X$?
 A: Let $R$ be any ring (commutative, with 1). Then $\text{End}_{R-\text{modules}}(R) = R$: any endomorphism $\phi$ is uniquely determined by $\phi(1)$ since $\phi(r) = r\phi(1)$, and $\phi(1)$ can be set to anything in $R$.
A: I don't know what you mean by "spaces," but here's a very general setup where you have an isomorphism of this sort. Consider a closed monoidal category, namely a category $M$ with a monoidal structure $\otimes$ and an internal hom $[-, -]$ fitting into an adjunction
$$\text{Hom}(a, [b, c]) \cong \text{Hom}(a \otimes b, c).$$
$\text{Vect}$ is such a category, with $\otimes$ the tensor product and $[-, -]$ the vector space of linear maps between two vector spaces. More generally, for $R$ a commutative ring, $\text{Mod}(R)$ is such a category, with $\otimes$ the tensor product over $R$ and $[-, -]$ the $R$-module of $R$-module homomorphisms between two $R$-modules. 

Proposition: The unit object $1$ always satisfies $1 \cong [1, 1]$. More generally, every object $b$ satisfies $b \cong [1, b]$.

Proof. Follows immediately from the adjunction $\text{Hom}(a, [1, b]) \cong \text{Hom}(a \otimes 1, b) \cong \text{Hom}(a, b)$ together with the Yoneda lemma. $\Box$
