Definition of monoidal categories in Etingof’s “Tensor Categories” I am quite confused on page 22,23 of the unit axiom of a monoidal category.

Unit axiom: $L_1:X \mapsto 1 \otimes X$ and $R_1:X \mapsto X \otimes 1$ are autoequivalences of $C$.
(end of pg 22) Choose natural isomorphisms $l_1:L_1 \Rightarrow \mathrm{id}$ such that it equals the diagram of compositions
$$1 \otimes (1\otimes X) \rightarrow (1 \otimes 1) \otimes X \rightarrow 1 \otimes X. $$

Why can we do this, and why is it a natural isomorphism?

This doesn't seem right. What I see here: given a functor $F:C \rightarrow D$, we may define $i:F \Rightarrow \mathrm{id}$, if there exists an isomorphism $i_X:F(X) \rightarrow X$ for each $X \in C$?
But this would not imply commutativity of the diagram. Since we require $i_Y F(f) = f \, i_X$ given a morphism $f:X \rightarrow Y$. So $F(f)$ would be determined, i.e. we cannot define a natural isomorphism from arbitrary $i_X$.
 A: The functor $(1\otimes-)\colon\mathcal{C}\to\mathcal{C}$ is an autoequivalence, hence the functor $(1\otimes-)^{\mathcal{C}}\colon\mathcal{C}^{\mathcal{C}}\to\mathcal{C}^{\mathcal{C}}$ is also an autoequivalence, where $\mathcal{C}^{\mathcal{C}}$ denotes the category of endofunctors of $\mathcal{C}$.
Hence $(1\otimes-)^{\mathcal{C}}$ is fully faithful and conservative. Thus, for every natural isomorphism $\gamma\colon(1\otimes(1\otimes-))\to(1\otimes-)$ there exists the unique natural isomorphism $\delta\colon(1\otimes-)\to I_{\mathcal{C}}$, such that $(1\otimes-)^{\mathcal{C}}(\delta)=\gamma$. So in your task you shoud only check that your $\gamma=(\imath\otimes I_{\mathcal{C}})\circ\alpha^{-1}(1,1,-)$ is a natural isomorphism. It follows from the following facts: $\alpha$ is a natural isomorphism, $\imath$ is an isomorphism, (horizontal and vertical) composition of natural isomorphisms is a natural isomorphism.
Appendix.
Definition 1. Let $\mathcal{A}$ and $\mathcal{B}$ be categories. An equivalence $4$-tuple between $\mathcal{A}$ and $\mathcal{B}$ is a $4$-tuple $(T,T',u_T,\varepsilon_T)$, such that $T\colon\mathcal{A}\to\mathcal{B}$ and $T'\colon\mathcal{B}\to\mathcal{A}$ are functors, $u_T\colon I_{\mathcal{A}}\to T'\circ T$ and $\varepsilon_T\colon T\circ T'\to I_{\mathcal{B}}$ are natural isomorphisms. In this case we say that $(T,T',u_T,\varepsilon_T)$ is an equivalence $4$-tuple for $T$.
Proposition 1. A functor $T$ is an equivalence iff there exists an equivalence $4$-tuple for $T$.
Proposition 2. Let $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$, $\mathcal{D}$ be categories, $T\colon\mathcal{A}\to\mathcal{C}$ and $S\colon\mathcal{D}\to\mathcal{B}$ be functors. Then if the functors $T$ and $S$ are equivalences, then the functor $T^S\colon\mathcal{A}^{\mathcal{B}}\to\mathcal{C}^{\mathcal{D}}$ is also an equivalence.
Proof. If $(T,T',u_T,\varepsilon_T)$ is an equivalence $4$-tuple for $T$ and $(S,S',u_S,\varepsilon_S)$ is an equivalence $4$-tuple for $S$, then $(T^S,T'^{S'},u_T^{\varepsilon_S^{-1}},\varepsilon_T^{u_S^{-1}})$ is an equivalence $4$-tuple for $T^S$.
Corollary 1. Let $\mathcal{A}$ be a category, $T\colon\mathcal{A}\to\mathcal{A}$ be an endofunctor on $\mathcal{A}$. Then if $T$ is an autoequivalence, then $T^{\mathcal{A}}\colon\mathcal{A}^{\mathcal{A}}\to\mathcal{A}^{\mathcal{A}}$ is also an autoequivalence.
Proof. By the Proposition 2 the functor $T^{\mathcal{A}}=T^{I_{\mathcal{A}}}$ is an equivalence.
