= Solution
Let $(A,a)$ be a $T$-algebra with at least two elements, and suppose $Ff:FX\to FY$ is an isomorphism. For every $T$-algebra $A$, precomposition with $Ff$ gives a bijection
$$
\mathbf{Set}^T(FY,A)\longrightarrow\mathbf{Set}^T(FX,A).
$$
By the free-forgetful adjunction this is
$$
\mathbf{Set}(Y,UA)\longrightarrow\mathbf{Set}(X,UA),
\qquad h\longmapsto hf.
$$
For one set $UA$ with at least two elements, bijectivity of this precomposition forces $f$ to be bijective: surjectivity detects injectivity of $f$, and injectivity detects surjectivity. Thus $f$ is an isomorphism. The free functor reflects isomorphisms, completing the cycle of equivalences.
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