If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself? If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself? 

For example the square of $2$  is $2^2=2 \cdot 2=4 $ .

But square root of $2$ is not $\frac{2}{2}=1$ .
taking square root means reversing the effect of squaring. Dividing a number by itself does not do that (but rather always returns 1 as you noted).

Compare your question to: if doubling a number means adding it to itself, shouldn't halving a number mean subtracting it from itself? Answer: obviously not.
If you take a number, $N$ and multiply it times $x$ and then again times $x$, you would end up with $Nx^2$. To get $N$ back, you would divide by $x$ and then divide again by $x$.

The word root is generally used to represent one of the zeros of a function; that is a solution to $f(x) = 0$. In particular, a square root of N  is a solution to the polynomial equation $x^2 - N = 0$. If $N$ is a non negative real number, then there is a positive square root and a negative square root of $N$. Usually we express the positive square root of $N$ as $\sqrt N$ and the negative square root of $N$ as $-\sqrt N$.

ALSO

I suppose that , if you want dividing by $x$ in the definition of square root, then you can say that the square root of $N$ means finding the number, $x$, such that $\dfrac Nx = x$.

A comparison

\begin{array}{ccc}
   \text{action} &amp; \text{equation} &amp; \text{inverse} \\
\hline
   \text{doubling} &amp; 2x = x + x &amp; x = \dfrac 12(2x) \\
   \text{squaring} &amp; x^2 = x \cdot x &amp; x = \sqrt{x^2} \\
   \hline
\end{array}

Notice that the inverse of doubling is not subtracting x and that the inverse of squaring is not dividing by x.