Let $T$ denote an algebraic theory. Then given a free $T$-algebra $F(k)$, the inclusion of generators map $\eta : k \rightarrow U(F(k))$ is usually injective, in practice. It doesn't have to be, though; consider the case where $T$ proves the identity $x \equiv y$. Then for $k \geq 2$, we have that $\eta : k \rightarrow U(F(k))$ is non-injective.
Question. What must we require of an algebraic theory $T$ for it to follow that for every cardinal number $k$, the inclusion of generators map $k \rightarrow U(F(k))$ is injective?
The inclusion of generators map $i_\kappa:\kappa\rightarrow U(F(\kappa))$ is injective if and only if $T$ does not imply the identity $x = y$.
You've noted one direction in your question. For the converse, note that if $T$ does not imply $x = y$, then $T$ has at least one model, $A$, with $|A|\geq 2$. Pick $a,b\in A$ with $a\neq b$. Then for any cardinal $\kappa$, for any $\alpha, \beta\in \kappa$ with $\alpha\neq \beta$, there is a map $f:F(\kappa)\rightarrow A$ such that for $U(f):U(F(\kappa))\rightarrow U(A)$, we have $i_\kappa(\alpha)\mapsto a$ and $i_\kappa(\beta)\mapsto b$. Since $a\neq b$, $i_\kappa(\alpha)\neq i_\kappa(\beta)$, so $i_\kappa$ is injective.