An explanation to the level of high-school mathematics might be a bit too hard, but let's try undergrad mathematics.
On a space of $d$ dimension (take $d=1,2,3$ if that's easier to visualize), namely $\mathbb{R}^d$, the Euclidean group can defined as the group of bijective transformations $T$ which preserve the Euclidean distance $|x-y|$, namely such that for all $x,y\in\mathbb{R}^d$, $|T(x)-T(y)|=|x-y|$. The group of (global) conformal transformations can be defined in a similar manner. First we need to add a fictitious "point at infinity" $\infty$, so we work with $\widehat{\mathbb{R}^d}=\mathbb{R}^d\cup\{\infty\}$ which has the same shape as a $d$-dimensional sphere.
For any quadruples $(x_1,x_2,x_3,x_4)$ of distinct elements of $\widehat{\mathbb{R}^d}$ define their absolute cross-ratio as
$$
C(x_1,x_2,x_3,x_4)=\frac{|x_1-x_3|\ |x_2-x_4|}{|x_1-x_4||x_2-x_3|}
$$
involving distances between certain pairs of points. If $|x_i-x_j|$ contains a point equal to $\infty$ then we set by definition $|x_i-x_j|=1$.
Now a conformal map $T$ is just a bijection from $\widehat{\mathbb{R}^d}$ to itself which conserves the cross ratio, i.e., such that
$$
C(T(x_1),T(x_2),T(x_3),T(x_4))=C(x_1,x_2,x_3,x_4)
$$
for all quadruples of distinct points.
Intuitively $T$, locally (near an arbitrary point say $x$) looks like a Euclidean isometry times a dilation by a factor $\lambda_x$.
Now a spin model (in $d=2$) like Ising can be thought of as a random black and white image (of infinite size). The pixels sit on a grid or lattice like $\mathbb{Z}^d$. An important question is to understand the large scale features of this random image. One way to do that is to to zoom out either by letting the observer move back with the image staying where it is, or equivalently (and preferably) by shrinking the image with the observer staying immobile. Since the image is of infinite size, it remains so during the shrinking operation. However, the pixels become smaller and smaller. If one can do this all the way to infinity, then the final result is a random image in the continuum (pixels have size zero now). Moreover, this has to be a random greyscale image rather than just black and white.
Statistical properties of this image can be made quantitative using so-called correlations
$$
\langle\phi(x_1)\cdots\phi(x_n)\rangle\ .
$$
For each point or location $x_i$ measure the amount of grey $\phi(x_i)$ in the particular image sample, then take the product, and finally average over the picture samples. Almost by definition, the random image in the continuum has scale invariant statistics, i.e.,
$$
\langle\phi(\lambda x_1)\cdots\phi(\lambda x_n)\rangle=\lambda^{-n\Delta}
\langle\phi(x_1)\cdots\phi(x_n)\rangle
$$
for all $n$, for all locations of the distinct points $x_1,\ldots,x_n$, and for all rescaling factors $\lambda>0$. Here $\Delta$ is a suitable constant called the scaling dimension of the random image.
If one believes there is some locality in how the statistics of the random picture are defined (e.g. short-range interactions between pixels only), then one could surmise that the above equation holds more generally if, instead of dilating everybody by the same amount $\lambda$, we do different dilations in different regions of space. This in a nutshell is conformal invariance.
Namely, we will say the random image is conformally invariant (we are dealing with a conformal field theory) if
$$
\langle\phi(T(x_1))\cdots\phi(T(x_n))\rangle=\lambda_{x_1}^{-\Delta}\cdots\lambda_{x_n}^{-\Delta}
\langle\phi(x_1)\cdots\phi(x_n)\rangle
$$
as above, now with $T$ being a conformal transformation. The local dilation factor $\lambda_x$ associated to $T$ can be computed as the $d$-th root of the change of volume factor. Since I am allowed undergrad math, i.e., multivariable calculus: this change of volume factor is given by the absolute value of the Jacobian determinant of $T$ at $x$.