Suppose we take a system of differential equations $\dot {\vec{x}}=\vec f(\vec x)$ and have a relation $\vec x(t_0+T)=\vec x(t_0)$ for some real $t_0$ and T when can we say that $\vec x(t+T)=\vec x(t)$ for all $t$ real and that particular T?
It seems to be true when $\vec x(t)$ is analytic(same taylor series expansion for all time remains valid) for all time t.
Can anyone suggest some other conditions when the relation in question is valid?
Since $x(t)=tan(t)$ has this property and satisfies the differential equation $\dot x =1+x^2$ with initial condition x(0)=0
Further this above system has periodicity in 1d. Am I doing something wrong here since there should not be periodicity in 1d autonomous systems?
Yes, this is true if $f$ is Lipschitz or differentiable. Then $x(t)$ and $y(t)=x(t+T)$ have the same values at $t_0$ and by the uniqueness theorem have to be equal everywhere.
Your counter example is not one as the solution $x(t)=\tan t$ is only defined on an interval between two neighboring poles. There is no continuation over a pole. If you set $x=-\frac{u'}{u}$, then the second order equation $u''+u=0$ or the first order system $(u',v')=(v,-u)$ will have periodic solutions as discussed in the first part. However, the solution $x$ of the original equation is only defined on one of the intervals between the roots of the denominator $u$.