Let $X$ be scheme over a field $k$ of characteristic $p>0$. Let $\mu_n$ denote the following sheaf in the fppf topology
$ \mu_p(T) = \{ t \in \Gamma(T, \mathcal{O_T^{*}}): t^p = 1 \} $.
Also let us define the sheaf $\alpha_p$ in the fppf topology
$ \alpha_p(T) = \{ t \in \Gamma(T, \mathcal{O_T}): t^p = 0 \} $.
Why is there no trivial sheaf homomorphism from $\mu_p$ to $\alpha_p$?
We may reduce to the case when $X$ is affine, say $X = \operatorname{Spec} A$ where $A$ is an $\mathbb{F}_{p}$-algebra. We have \begin{align} \mu_{p} &= \mathrm{Mor}_{\mathrm{Sch}/A}(-,\operatorname{Spec} A[s]/(s^{p}-1)) \\ \alpha_{p} &= \mathrm{Mor}_{\mathrm{Sch}/A}(-,\operatorname{Spec} A[t]/(t^{p})) \end{align} and a morphism of group sheaves $$ \mu_{p} \to \alpha_{p} $$ corresponds (via the Yoneda lemma) to a morphism of $A$-schemes $$ \operatorname{Spec} A[s]/(s^{p}-1) \to \operatorname{Spec} A[t]/(t^{p}) $$ that respects the group laws. In turn this corresponds to an $A$-algebra homomorphism $$ A[s]/(s^{p}-1) \leftarrow A[t]/(t^{p}) $$ that respects the coalgebra structure. Let $$ f(s) \in A[s]/(s^{p}-1) $$ be the image of $t$ under the $A$-algebra map, viewed as an $A$-linear combination of $1,s,\dotsc,s^{p-1}$. Then the condition about respecting the coalgebra structure amounts to saying that the diagram $\require{AMScd}$ \begin{CD} A[s]/(s^{p}-1) @<{\xi_{1}}<< A[t]/(t^{p}) \\ @V{\xi_{3}}VV @VV{\xi_{4}}V\\ A[s_{1},s_{2}]/(s_{1}^{p}-1 , s_{2}^{p}-1) @<<{\xi_{2}}< A[t_{1},t_{2}]/(t_{1}^{p},t_{2}^{p}) \end{CD} commutes, where $\xi_{1}$ sends $t \mapsto f(s)$ and $\xi_{2}$ sends $(t_{1},t_{2}) \mapsto (f(s_{1}),f(s_{2}))$ and $\xi_{3}$ sends $s \mapsto s_{1}s_{2}$ and $\xi_{4}$ sends $t \mapsto t_{1}+t_{2}$. Thus commutativity of the above diagram corresponds to the condition that $$ f(s_{1}s_{2}) = f(s_{1}) + f(s_{2}) $$ in $A[s_{1},s_{2}]/(s_{1}^{p}-1 , s_{2}^{p}-1)$, where this ring is a finite free $A$-module of rank $p^{2}$, with basis $s_{1}^{e_{1}}s_{2}^{e_{2}}$ for $0 \le s_{1},s_{2} \le p-1$. The "support" of the term $f(s_{1}s_{2})$ consists of monomials of the form $(s_{1}s_{2})^{e}$, whereas the "support" of the term $f(s_{1}) + f(s_{2})$ consists of monomials of the form $s_{1}^{e}$ or $s_{2}^{e}$, and their "intersection" is precisely $1 = s_{1}^{0} s_{2}^{0}$; thus $f$ must be constant. Then $f = f + f$ means $f = 0$.