Given (non-negative) $x$ and $y$, where their length is the same (i.e., |x|=|y|): if (1) mean of $x$ and mean of $y$ are the same (i.e., $\overline{x}=\overline{y}$); (2) their standard deviations are the same (i.e., $\sigma_x=\sigma_y$); and (3) product of standard deviations of $x$ and $y$ is equal to covariance of $x$ and $y$ (i.e., $\sigma_x\sigma_y =\sigma_{xy}$); does this mean $x=y$?
I know $\sigma_x\sigma_y =\sigma_{xy}$ is the equality version of Cauchy-Schwarz, and that this one holds iff $y=ax+b$ (see Equality in Cauchy-Schwarz); that is: if $x$ is a scalar multiple of $y$ (or vice versa). But I do not know if I can use this fact for the proof.
Can anyone provide a simple proof? I am not a mathematician.
Yes - under those conditions it must be that $x = y$. In fact, this must be true even if you don't know beforehand that $|x| = |y|$.
You are mixing up some concepts here, while dropping key components behind those concepts. "mean", "standard deviation", and "covariance" are concepts about collections of data. "Norm", "inner product", "Cauchy-Schwarz inequality" are concerned with vectors.
If you treat your data sets as the coordinates of vectors ($x$ and $y$ will have to be the same size for this), then these two areas overlap, and you can compare the concepts in each. But be sure you know how they translate to each other. standard deviation and covariance do look very similar to the norm and inner product, but they are defined relative to the mean, while the inner product and norm are relative to the origin of the vector space. So if you want to equate these concepts, you cannot just treat your data set directly as a vector. Instead, you need to subtract out the mean. Denoting the dataset $x$ as "$[x]$" and the vector $x$ as "$\vec x$" $$\vec x = [x] - \bar x = (x_1 - \bar x, x_2 - \bar x, \dots, x_n - \bar x)$$ And similarly for $\vec y$. With this translation, we find that $$\|\vec x\| = \sqrt n\sigma_x\\\|\vec y\| = \sqrt n\sigma_y\\\langle \vec x, \vec y\rangle = n\sigma_{xy}$$ And so if $\sigma_x\sigma_y = |\sigma_{yx}|$, it follows that $\|\vec x\|\|\vec y\| = |\langle \vec x, \vec y\rangle|$, which by Cauchy-Schwarz is true if and only $\vec y = a\vec x$ for some $a$.
But then, we also have $\sigma_x = \sigma_y$, which gives $$\|\vec x\| = \|\vec y\| = \|a\vec x\| = |a|\|\vec x\|$$ Assuming $\|\vec x\| \ne 0$, we must have $|a| = 1$, or $a = \pm 1$, and therefore $\vec y = \pm \vec x$. But if $\vec y = -\vec x$ then $$\langle \vec x, \vec y\rangle = -\langle \vec x, \vec x\rangle = -\|\vec x\|^2 = -\|\vec x\|\|\vec y\|\\\sigma_{xy} = -\sigma_x\sigma_y$$ which we know to be false. Thus $\vec y = \vec x$ is the only possibility, or translating back to the original problem, $y = x$.
Note that nowhere was $|x| = |y|$ used, so that assumption was unnecessary.