TITLE: Quick proof that free objects on sets of different cardinality are not isomorphic?
QUESTION [6 upvotes]: Suppose $F\dashv U$ is a free-forgetful adjunction of some algebraic theory. I thought there should be a one line proof that if $X\not\cong Y$ then $FX\not\cong FY$, i.e free objects on sets of different cardinality are not isomorphic. I am nowhere near finding one.
Having poked around on MSE, I don't see any hint of such a proof. The case of groups uses a lot of special structure which doesn't seem applicable to general algebraic theories. Now I also kind of doubt whether my intuitive claim is even true.
Is the claim true? Is there a (slick) proof?

REPLY [3 votes]: The standard counterexample is to take a take $R=\mathrm{End}(V)$ for some infinite-dimensional vector space $V$. Then $R \cong R^2$ as left $R$-modules.