= Solution
Use letters $x_i$ and their formal inverses. An elementary contraction deletes an adjacent pair $x_ix_i^{-1}$ or $x_i^{-1}x_i$. A <reduced word> has no such pair. Repeated contraction terminates because each step reduces length by two, so every word has a reduced descendant.
For uniqueness, two possible first contractions either occur on disjoint pairs, when performing both yields a common descendant, or overlap in three letters $aa^{-1}a$, when either contraction produces the same remaining letter. Induct on word length: the shorter descendants of both first contractions therefore have the same eventual <reduced word>. This proves every contraction sequence from a given word ends at the same <reduced word>. An insertion or deletion of an inverse pair preserves that reduced result. Since the word model of the <free group> identifies words exactly under finite sequences of these moves, every <group> element has \b[a unique reduced representative].
To define the requested <homomorphism>, substitute $w_i$ for $x_i$ and $w_i^{-1}$ for $x_i^{-1}$ in any word. Inverse-pair insertions or deletions map to inverse words that cancel, so substitution is well defined on <group> elements and respects concatenation. This proves the universal property of the <free group> in this case.
For an <automorphism> of order four, take $\phi(x_1)=x_2$ and $\phi(x_2)=x_1^{-1}$. Its iterates on $x_1$ are $x_2,x_1^{-1},x_2^{-1},x_1$; its fourth iterate also fixes $x_2$. Unique reduced representatives show the first three iterates on $x_1$ differ from $x_1$. Hence \b[its order is exactly four].
For the final map, every iterate of $x_2$ is a positive word and contains an $x_2$. Substituting once replaces each occurrence of $x_2$ by $x_2w$ and leaves each $x_1$ unchanged. If $\ell=|w|\ge1$, its length consequently increases by at least $\ell$ on every iteration. No cancellation occurs in a positive word, so $|\phi^n(x_2)|\ge1+n\ell>1$. It cannot equal $x_2$, proving \b[no positive iterate of this <endomorphism> is the identity]. Surjectivity of this last <endomorphism> is not assumed.
This uses <positive-word growth under a free-group endomorphism>.
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