Graphs can look differently but be identical. The only thing that matters are which nodes are connected to which other nodes. If there exists an edge-preserving bijection between two graphs, in other words if there exists a function that maps nodes from one graph onto those of another such that the set of connections for each node remain identical, the two graph are said to be isomorphic.
\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}[
vertex/.style = {circle, draw, inner sep=1pt, fill=white},
vertex1/.style = {vertex, fill=red!30!white},
vertex2/.style = {vertex, fill=orange!30!white},
vertex3/.style = {vertex, fill=blue!30!white},
vertex4/.style = {vertex, fill=teal!30!white},
]
\draw[thick]
(0,0) node[vertex1] (n1^1) {$n_1$}
-- (0,2) node[vertex2] (n2^1) {$n_2$}
-- (2,2) node[vertex3] (n3^1) {$n_3$}
-- (2,0) node[vertex4] (n4^1) {$n_4$} -- cycle;
\begin{scope}[xshift=4cm]
\draw[thick]
(0,0) node[vertex1] (n1^2) {$n_1$}
-- (2,2) node[vertex2] (n2^2) {$n_2$}
-- (0,2) node[vertex3] (n3^2) {$n_3$}
-- (2,0) node[vertex4] (n4^2) {$n_4$} -- cycle;
\end{scope}
\begin{scope}[xshift=8cm]
\draw[thick]
(0,0) node[vertex1] (n1^3) {$n_1$}
-- (2,2) node[vertex2] (n2^3) {$n_2$}
-- (2,0) node[vertex4] (n3^3) {$n_3$}
-- (0,2) node[vertex3] (n4^3) {$n_4$} -- cycle;
\end{scope}
\begin{scope}[xshift=12.5cm]
\draw[thick]
(-0.5,0) node[vertex1] (n1^4) {$n_1$}
-- (0.25,2.2) node[vertex2] (n2^4) {$n_2$}
-- (2,1.6) node[vertex3] (n3^4) {$n_3$}
-- (-0.7,1.4) node[vertex4] (n4^4) {$n_4$} -- cycle;
\end{scope}
\end{tikzpicture}
\end{document}