$X^X$ as a monoid in a closed monoidal category $\require{AMScd}$
Consider a symmetric monoidal category $(\mathcal{C}, \otimes)$ with an adjunction $(A\otimes B, C) \simeq (A, C^{B})$ id est an internal hom set functor.
For every object $X$ there is a canonical arrow $\begin{CD}X^X\otimes X^X @>{m}>>X^X\end{CD}$ given as the adjunct of $\begin{CD}X^X \otimes X^X \otimes X @>{X^X \otimes \: \text{ev}}>> X^X \otimes X @>{\text{ev}}>> X \end{CD}$ where $\text{ev}$ is the evaluation map.
In $\mathbf{Set}$ this arrow gives $X^X$ a monoid structure. I'm wondering if it is true in the general case.
I tried to prove it by drawing a lot of diagrams, but failed, probably from want of a global intuition.
A related question (though the first one seems more important) is to prove that $X\otimes X \overset{f}{\rightarrow} X$ defines an associative law if and only if the diagram
$\begin{CD}
X\otimes X@>{f^{adj}}>> X^X \otimes X^X \\
@VV{f}V @VV{m}V \\
X @>{f^{adj}}>> X^{X}
\end{CD}$
commutes. In $\mathbf{Set}$ it means that multiplication by $a$ followed by multiplication by $b$ is the same as multiplication by $ba$.
All of these questions arised when trying to prove both definitions of $\mathbb{N}$ as an universal pointed arrow or an universal pointed monoid are equivalent.
 A: Yeah, this is definitely a monoid. The main thing to prove is associativity, i.e. that $m\circ(X^X\otimes m)=m\circ(m\otimes X^X)$. The way to proceed is to show that their adjuncts are equal.
The adjunct of $m\circ(X^X\otimes m)$ is
$$\begin{CD}X^X \otimes X^X \otimes X^X \otimes X @>{X^X \otimes\,m\,\otimes X^X}>>X^X \otimes X^X \otimes X @>{X^X \otimes \: \text{ev}}>> X^X \otimes X @>{\text{ev}}>> X \end{CD}$$
which we can rewrite as
$$\begin{CD}X^X \otimes X^X \otimes X^X \otimes X @>{X^X \otimes X^X\otimes \mathrm{ev}}>>X^X \otimes X^X \otimes X @>{\text{ev}\circ(m\otimes X)}>> X \end{CD}$$
and the adjunct of $m\circ(m\otimes X^X)$ is
$$\begin{CD}X^X \otimes X^X \otimes X^X \otimes X @>{m\,\otimes X^X \otimes X^X}>>X^X \otimes X^X \otimes X @>{X^X \otimes \: \text{ev}}>> X^X \otimes X @>{\text{ev}}>> X \end{CD}$$
which we can rewrite as
$$\begin{CD}X^X \otimes X^X \otimes X^X \otimes X @>{X^X \otimes(\text{ev}\circ(m\otimes X))}>>X^X \otimes X @>{\text{ev}}>> X \end{CD}.$$
So our trick will be to show that $\text{ev}\circ(m\otimes X)=\text{ev}\circ(X^X\otimes \text{ev})$ since then the adjuncts of both $m\circ(X^X\otimes m)$ and $m\circ(m\otimes X^X)$ will be equal to
$$\begin{CD}X^X \otimes X^X \otimes X^X \otimes X @>{X^X \otimes X^X \otimes \: \text{ev}}>>X^X \otimes X^X \otimes X @>{X^X \otimes \: \text{ev}}>> X^X \otimes X @>{\text{ev}}>> X \end{CD}.$$
But it's immediate that $\text{ev}\circ(m\otimes X)=\text{ev}\circ(X^X\otimes \text{ev})$ since the adjunct (going the other way now) of $\text{ev}\circ(m\otimes X)$ is $m$ (by definition of $\text{ev}$) and the adjunct of $\text{ev}\circ(X^X\otimes \text{ev})$ is also $m$ (by definition of $m$).
A similar argument works to show the identity law, where the identity is defined as the adjunct of $\lambda_X:I\otimes X\to X$.
