Suppose $T$ is a linear operator $(V,V)$, and $T^2=T$. Does this means it is the identity operator? Since we're saying $T^2=T$, isn't $T$ a square root of itself? So that would only happen if $T=I$, correct? 
Another thing I tried is, for $u$, $v$ in $V$, where $Tu=v, TTu=Tu \to T(v)=v  \to T=I$
is this correct?

Both zero and the identity satisfy this. More generally, any linear operator which is diagonalizable and has only $1$s,  and $0$s as eigenvalues satisfies this equation.
So the answer is no.
