Consider $\Bbb F_{p^n}$ as a vector space over $\Bbb F_{p}$. Now let $T$ be a non-zero $\Bbb F_{p}$-linear map from $\Bbb F_{p^n}$ to $\Bbb F_{p^n}$. Now if $0$ and $\Bbb F_{p^n}$ are the only $T$-invariant subspace of $\Bbb F_{p^n}$ then we have to prove that $T=aI$ (i.e multiplication by $a$) where $\Bbb F_{p^n}=\Bbb F_{p}(a)$ (i.e $a$ is a primitive element of $\Bbb F_{p^n}$ over $\Bbb F_{p}$).
What I have tried:
Clearly $T$ is an isomorphism. So $T$ has finite order as an element in the invertible matrix group over $\Bbb F_{p}$, say $d$. Then for every non-zero element $v$, $v+T(v)+T^2(v)+\dots+T^{d-1}(v)=0$.
But I can't proceed further. I need some help.
Taken literally, the claim is FALSE as stated. Consider the case $p=3$, $n=2$. We construct the field $\Bbb{F}_9$ as $\Bbb{F}_3[x]/\langle x^2+1\rangle$. Let $\alpha$ denote a root of $x^2+1$ in $\Bbb{F}_9$. Let's use the basis $1,\beta=1+\alpha$, and define the mapping $T$ by $T(1)=1+\alpha$, $T(1+\alpha)=-1$, i.e. $T$ is the linear transformation with the matrix $$ M(T)=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right). $$ There are no non-trivial $T$-invariant subspaces for such a subspace would be 1-dimensional over $\Bbb{F}_3$ and thus an eigenspace of $T$ belonging to eigenvalue $\in\Bbb{F}_3$. But the eigenvalues of $T$ are fourth roots of unity. Also, $T$ is not multiplication by a generating element $a$ of $\Bbb{F}_9$. As $T(1)=1+\alpha$ the only possibility would be $a=1+\alpha$. But $T(1+\alpha)=-1\neq(1+\alpha)^2=2\alpha$.
The problem with the above "literal" interpretation of the question is that you began by declaring the structure of a field to $\Bbb{F}_p^n$. A true statement would be as follows.
Claim. Let $T:\Bbb{F}_p^n\to\Bbb{F}_p^n$ be a linear mapping that has no non-trivial invariant subspaces. Then the ring $R=\Bbb{F}_p[T]$ gotten by adjoining the powers of $T$ to the field of scalar matrices is a field of $p^n$ elements, IOW $R\cong \Bbb{F}_{p^n}$.
Proof. Consider the minimal polynomial $p(x)\in\Bbb{F}_p[x]$ of $T$. If it is of degree less than $n$, then we easily get a non-trivial $T$-invariant subspace. The same holds if $p(x)$ is not irreducible (the details depend a little on whether $p(x)$ has a factor with multiplicity $>1$ or whether it's separable). This implies that the subring $R$ of $M_n(\Bbb{F}_p)$ generated by $T$ and the scalar matrices is isomorphic to $\Bbb{F}_p[x]/\langle p(x)\rangle\cong\Bbb{F}_{p^n}$. Q.E.D.
The moral. We can embed the field $\Bbb{F}_{p^n}$ into $M_n(\Bbb{F}_p)$ in many somewhat unrelated ways. I say "somewhat unrelated", because if $\phi_1$ and $\phi_2$ are two such embeddings, then by the Skolem-Noether Theorem they differ from each other by conjugation. More precisely, there exists an invertible matrix $A\in GL_n(\Bbb{F}_p)$ such that $$ \phi_2(x)=A\phi_1(x)A^{-1} $$ for all $x\in\Bbb{F}_{p^n}$.