Classifying the groups of order 56
A preliminary remark (which I think you understand): there is exactly one way to make the group $(\mathbb{Z}_p)^n=\mathbb{Z}_p \oplus\cdots\oplus \mathbb{Z}_p $ (I write it additively) into a vector space over the field of $p$ elements $\mathbb{F}_p$: for the axioms to be true you must set $k\cdot x=x+x+\cdots+x$, the sum of $k$ $x$'s. This means that the group automorphisms are also vector space automorphisms, and so the automorphism group is exactly $\text{GL}(n,\mathbb{F}_p)$.
I think that if you want to tackle questions like this you do need to know something about Linear Algebra: the theorem of central importance here is (I suggest) the Rational Canonical Form Theorem. I think it is a good idea to understand its use first, before worrying about eigenvalues: that often involves extending the field (as in @Derek Holt 's comment) and that is not quite so intuitive.
In the case you are considering you want to find an $\alpha:V_3(\mathbb{F}_2)\to V_3(\mathbb{F}_2)$ which is of order $7$.
You know, then, that $\alpha^7=1$, and so the minimal polynomial $m_{\alpha}(X)$ of $\alpha$ must divide $X^7-1$.
Over $\mathbb{F}_2$ we can calculate easily that $X^7-1=(X-1)(X^3+X^2+1)(X^3+X+1)$, and both the cubic factors must be irreducible.
Disregarding the trivial case we must have that $m_{\alpha}(X)=X^3+X^2+1$ or $X^3+X+1$. The Theorem of the Rational Canonical Form will now tell us that every $\alpha$ is conjugate to the companion matrix of one or other of these cubics. That is we may, by choosing a suitable basis of $V_3(\mathbb{F}_2)$, assume that $\alpha$ is one of:
$$ \begin{pmatrix} 0 & 0 & 1\\ 1& 0& 0\\ 0 & 1 & 1\\ \end{pmatrix} \text{ or } \begin{pmatrix} 0 & 0 & 1\\ 1& 0& 1\\ 0 & 1 & 0\\ \end{pmatrix}. $$
Note that if $\alpha$ satisfies $X^3+X^2+1$ then $\alpha^{-1}$ satisfies the other irreducible, $X^3+X+1$. So in considering non-trivial group extensions of the elementary abelian group of order $8$ by a cyclic group of order $7$ there is really only one possibility: $$ \langle x,y,z,s |x^2=y^2=z^2=[x,y]=[x,z]=[y,z]=1, s^7=1, x^s=y, y^s=z, z^s=xz \rangle. $$
(Exercises for the reader: find all groups of order $80$ with an elementary abelian normal subgroup of order $16$, and all groups of order 351 with an elementary abelian subgroup of order $27$.)
Viewing the 3-dimensional vector space over $\mathbb{F}_2$ as the field extension $\mathbb{F}_2\leq \mathbb{F}_8$ is very advantegous in this problem. You are looking for a matrix that is a root of the polynomial $x^7-1$, or to make the connection more apparent, a root of the polynomial $x^8-x$. Roots of this polynomial are exactly the elements of $\mathbb{F}_8$. So Factorize the polynomial into irreducible components:
$x^8-x= x(x-1)(x^3+x+1)(x^3+x^2+1)$. This is easy to find: we know that we are looking for two irreducible polynomials of degree 1 and two of degree 3 by general theory of finite fields.
So an easy way to construct a $3\times 3$ matrix over $\mathbb{F}_2$ of order 7 is to produce the companion matrix of the polynomial $x^3+x+1$, for example.
So one solution is: $\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$
Clearly, there is exactly one non-Abelian group of order 56 with an elementary Abelian normal subgroup of order 8. As you pointed out, we are looking for a nontrivial homomorphism $\varphi: \mathbb{Z}_7\rightarrow GL(3,2)$. You can construct such a homomorphism using the above matrix, for example, so existence is clear. As for uniqueness, you need two observations.
If two such homomorphisms have the same image, say $Im(\varphi)=Im(\psi)$, then one can be obtained from the other by composition with an automorphims from the right: $\varphi=\psi\circ \alpha$, $\alpha\in Aut(\mathbb{Z}_7)$.
All 7-element subgroups of $GL(3,2)$ are conjugate. So given $\varphi, \psi$, you can make their images coincide by composition with an automorphism from the left: $Im(\varphi)= Im(\beta\circ \psi)$, $\beta\in Aut(GL(3,2))$.
Composition with automorphisms yield isomorphic semidirect products.