Lemma: Let us work over a field ${\mathbf{k}}$. Let $p_1,\ldots,p_5,q_1,\ldots,q_5\in \mathbb{P}^2=\mathbb{P}^2_{\mathbf{k}}$ be ten points such that no $3$ of the $p_i$ are collinear and no $3$ of the $q_i$ are collinear (in particular, $\{p_1,\ldots,p_5\}$ consists of $5$ points and the same holds for $\{q_1,\ldots,q_5\}$). Then, the following are equivalent:
$a)$ There exists $\alpha\in \mathrm{Aut}(\mathbb{P}^2)=\mathrm{PGL}(3,{\mathbf{k}})$ such that $\alpha(\{p_1,\ldots,p_5\})=\{q_1,\ldots,q_5\}$.
$b)$ Denoting by $C\subset \mathbb{P}^2$ the unique irreducible conic through the $p_i$ and by $C'\subset \mathbb{P}^2$ the unique irreducible conic through the $q_i$, there is an isomorphism $C\to C'$ sending $\{p_1,\ldots,p_5\}$ onto $\{q_1,\ldots,q_5\}$.
$c)$ The Del Pezzo surfaces $X$ and $X'$ obtained by the blowing-ups $\pi\colon X\to \mathbb{P}^2$ of the $p_i$ and $\pi'\colon X'\to \mathbb{P}^2$ of the $q_i$ respectively are isomorphic.
Remark: Condition $b)$ is in practice easy to check, as it is only a question on $5$ points of $\mathbb{P}^1$; one can for instance use the cross-ratios of $4$ of the $5$ points. In particular, over $\mathbf{k}=\mathbb{C}$ (or more generally when $\mathbf{k}$ is infinite), not moving $p_1,\ldots,p_4$ and changing $p_5$ gives infinitely many distinct non-isomorphic del Pezzo surfaces.
Proof of the Lemma:
The equivalence between $a)$ and $b)$ follows from the uniqueness of the conic through five points (which follows from the hypothesis that no $3$ are collinear).
$a)\Rightarrow c)$ follows from the universal property of blow-ups.
Hence, the only non-trivial (which if our interest here) is $c)\Rightarrow a)$.
Let $\tau\colon X\to X'$ be an isomorphism and consider the birational map $\varphi=\pi'\circ \tau\circ\pi^{-1}\mathbb{P}^2\dashrightarrow \mathbb{P}^2$. It has base-points only at the points $p_1,\ldots,p_n$. The preimage of a line is a curve of degree $d\ge 1$ having multiplicity $m_i$ at each $p_i$. Computing the self-intersection and using the adjunction formula on $X$, we get the two so-called Noether equalities
$$\sum m_i^2=d^2-1, \sum m_i=3d-1$$
and the only solutions are then
$(i)$: $d=1$, $(m_1,\ldots,m_5)=(0,0,0,0,0)$.
$(ii)$: $d=2$, $(m_1,\ldots,m_5)=(1,1,1,0,0)$ (up to permutation)
$(iii)$: $d=3$, $(m_1,\ldots,m_5)=(2,1,1,1,1)$ (up to permutation)
In case $(i)$, the map $\varphi$ is an automorphism of $\mathbb{P}^2$ which sends $\{p_1,\ldots,p_5\}$ onto $\{q_1,\ldots,q_5\}$, so we obtain $a)$.
In case $(ii)$, we change the coordinates such that $p_1=[1:0:0]$, $p_2=[0:1:0]$, $p_3=[0:0:1]$, $p_4=[1:1:1]$, $p_5=[a:b:c]$ with $a,b,c\in {\mathbf{k}}^*$. Then, $\varphi$ is equal to $\alpha\circ \sigma$, where $\alpha\in \mathrm{Aut}(\mathbb{P}^2)$ and $\sigma$ is the involution
$$[x:y:z]\mapsto [ayz:bxz:cxy]$$
As the points $p_4,p_5$ are exchanged by $\sigma$ and as $p_1,p_2,p_3$ are also base-points of the inverse of $\sigma$, we obtain $\alpha(\{p_1,\ldots,p_5\})=\{q_1,\ldots,q_5\}$.
In case $(iii)$, we again change the coordinates such that $p_1=[1:0:0]$, $p_2=[0:1:0]$, $p_3=[0:0:1]$, $p_4=[1:1:1]$, $p_5=[a:b:c]$ as before, and use the same $\sigma$ as before. Then, $\sigma\pi\colon X\to \mathbb{P}^2$ is again a birational morphism, contracting $5$ $(-1)$-curves onto the five points $p_1,\ldots,p_5$, and the birational map $\varphi\circ \sigma$ has now degree $2$. We again obtain $a)$ by using a second time a quadratic map as in $(ii)$.
