Let $p$ be an odd prime. For a non-abelian group $G$ of order $p^3$ in which no element is of order $p^2$, it is isomorphic to a semidirect product $$(\mathbb{F}_p\times\mathbb{F}_p)\rtimes\mathbb{F}_p.$$ Here I am trying to see if such semidirect product is unique up to isomorphism.
Note that $$\operatorname{Aut}(\mathbb{F}_p\times\mathbb{F}_p)\cong\operatorname{GL}(2,\mathbb{F}_p),$$ so for convenience, we shall consider a nontrivial group homomorphism $\phi:\mathbb{F}_p\to\operatorname{GL}(2,\mathbb{F}_p)$.
Since $\phi$ is nontrivial, it must be injective in this case, so $\phi(1)=:A$ is of order $p$. It has been shown from this post that $A$ has Jordan canonical form $J:=J_2(1)=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. That is, $A=BJB^{-1}$ for some $B\in\operatorname{GL}(2,\mathbb{F}_p)$.
Now let $\psi:\mathbb{F}_p\to\operatorname{GL}(2,\mathbb{F}_p)$ be another nontrivial group homomorphism. Then $\psi(1)=CJC^{-1}$ for some $C\in\operatorname{GL}(2,\mathbb{F}_p)$. By the following theorem:
Theorem. Let $H$ and $K$ be two groups and $\phi,\psi:K\to\operatorname{Aut}(H)$ be group homomorphisms. If $\psi=\phi\circ f$ for some $f\in\operatorname{Aut}(K)$, then $$H\rtimes_\phi K\cong H\rtimes_\psi K.$$
it suffices to find some $f\in\operatorname{Aut}(\mathbb{F}_p)$ such that $\psi=\phi\circ f$. Suppose $f(1)=k$. Then $$(\phi\circ f)(1)=\phi(f(1))=\phi(k)=\phi(1)^k=(BJB^{-1})^k=BJ^kB^{-1}.$$ If $\psi=\phi\circ f$, then we shall have $$CJC^{-1}=\psi(1)=(\phi\circ f)(1)=BJ^kB^{-1}\implies (B^{-1}C)J=J^k(B^{-1}C).$$ Let $S:=B^{-1}C=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$. It follows that $$\begin{bmatrix} a & a+b \\ c & c+d \end{bmatrix}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=SJ=J^kS=\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} a+kc & b+kd \\ c & d \end{bmatrix}.$$ Then we have $c=0$ and $a=kd$, which are not necessarily satisfied since $B$ and $C$ are arbitrary. Then I got stuck with this step.
I'd appreciate it if anyone has good ideas on this.
So far, you have proven that $\phi$ and $\psi$ differ by $\phi(t)=M^{-1} \psi(t) M$, where $M \in \operatorname{Aut}(\mathbb{F}^p \times \mathbb{F}^p)$. (Here $M=CB^{-1}$ if I'm not mistaken, but we will not need the explicit expression.) Now, you can use this to construct a explicit isomorphism. I imagine that this is similar in spirit to the theorem you quoted anyway.
Consider $$ \Phi\colon H \rtimes_\phi K \to H \rtimes_\psi K, \qquad \Phi(h, k)=(Mh, k). $$ This is of course bijective, so we only have to check that the map is a group homomorphism. In fact, we have $$ \Phi(h_1, k_1) \cdot \Phi(h_2, k_2)=(Mh_1, k_1) \cdot (Mh_2, k_2)=(Mh_1 \cdot \psi(k_1)(Mh_2), k_1\cdot k_2), $$ and $$ \Phi((h_1, k_1) \cdot (h_2, k_2))= \Psi(h_1 \cdot\phi(k_1)(h_2), k_1\cdot k_2)= (Mh_1 \cdot M \phi(k_1)(h_2), k_1 \cdot k_2). $$ But these expressions are equal, as $M\phi(t)=\psi(t)M$.