The identity functor $1_{\mathscr C}:\mathscr C \to \mathscr C$ has $1_{\mathscr C}(a)=a$, $1_{\mathscr C}(f)=f$. Verify that it is indeed a functor, specifically that it is a function that assigns:
To each $\mathscr C$-object $a$, a $\mathscr C$-object $1_{\mathscr C}(a)$
To each $\mathscr C$-arrow $f:a\to b$ a $\mathscr C$-arrow $1_{\mathscr C}(f):1_{\mathscr C}(a)\to 1_{\mathscr C}(b)$ such that
$1_{\mathscr C}(1_a)=1_{1_{\mathscr C}(a)}$ for all $\mathscr C$-objects $a$
$1_{\mathscr C}(g\circ f)=1_{\mathscr C}(g)\circ 1_{\mathscr C}(f)$, whenever $g\circ f$ is defined
2i is where I do not understand. LHS is supposedly an arrow (because it is a functor applied to an arrow), while RHS is supposed to be an object (because it is an arrow applied to a functor applied to $a$) - so how can an arrow be identical to an object?
From the LHS we have $1_{\mathscr C}(1_a)\to 1_a$ by def. that $1_{\mathscr C}(f)=f$.
From the RHS, we have $1_{1_{\mathscr C}(a)}\to 1_a$ by def. that $1_{\mathscr C}(a)=a$. This looks identical to what I have for LHS above...but I don't actually know what this is - is this supposed to be an arrow now? How did we go from dealing with an object to an arrow?
To avoid confusion it might be better to use another notation for identity arrows here.
Doing so 2i states that:$$1_{\mathscr C}(\mathsf{id}_a)=\mathsf{id}_{1_{\mathscr C}(a)}$$
Note that both LHS and RHS are (identity) arrows.