Let $R$ and $S$ be rings. If $h : R \to S$ is a ring homomomorphism, then by setting $$ \overline h\left( a_n x^n + \ldots + a_1 x + a_0 \right) := h(a_n) x^n + \ldots + h(a_1) x + h(a_0) $$ we have a homomorphism $\overline h : R[X] \to S[X]$. If $\operatorname{ev}_{R,\alpha} : R[X] \to R$ and $\operatorname{ev}_{S, \beta} : S[X] \to S$ denote the evaluation homomorphism at $\alpha \in R$ and $\beta \in S$ we have for $p \in R[X]$ $$ h(\operatorname{ev}_{R, \alpha}(p)) = \operatorname{ev}_{S, h(\alpha)}(\overline h(p)) $$ with the above constructions.
Are there any conditions that guarantee that every homomorphism $\overline h : R[X] \to S[X]$ determines a homomorphism $h : R \to S$ canonically? The choice $$ h(r) := \operatorname{ev}_{S, \beta}(\overline h(r)) $$ for some $\beta \in S$ gives a homomorphism, but it is not canonical.
Note: Just a subtle point in the above definition. We identify elements $r, r' \in R$ with their constant polynomials, and interpret $r\cdot r'$ as a polynomial product. Hence $\overline h(r\cdot r') = \overline h(r) \cdot \overline h(r')$.
Of course not. If $h\colon R\to S$ is a ring homomorphism, then $\bar{h}\colon R[X]\to S[X]$ satisfies $$ \bar{h}(X)=X $$ So a ring homomorphism $f\colon R[X]\to S[X]$ such that $f(X)\ne X$ cannot be of the form $\bar{h}$, for any $h\colon R\to S$.
Example: $f\colon R[X]\to R[X]$ which is the identity on $R$ and $f(X)=1+X$.
More generally, every pair $(h,\alpha)$, where $h\colon R\to S[X]$ is a ring homomorphism and $\alpha\in S[X]$ defines a ring homomorphism $h_\alpha\colon R[X]\to S[X]$ by stating that $h_\alpha(r)=h(r)$, for $r\in R$ and $h_\alpha(X)=\alpha$. This is actually a complete classification of the ring homomorphisms $R[X]\to S[X]$.