Are CDH and Square-DH equally hard? CDH problem roughly says that choose $U=g^u, V=g^v$ uniformly at random from  cyclic group $G$, it's hard to compute $\operatorname{CDH}(U,V)=g^{uv}$.

Square-DH problem roughly says choose $U=g^u$ uniformly at random from  cyclic group $G$, it's hard to compute $Z=g^{u^2}$

If I can solve the CDH problem, then it's very clear that Square-DH problem can also be solved easily: $\operatorname{CDH}(U,U)=g^{u^2}$. While, if Square-DH problem can be solved, then we can solve CDH problem of $UV$:
 $$g^{{(u+v)^2}}=g^{u^2+v^2+2uv}=g^{u^2}g^{v^2}\operatorname{CDH}(U,V)^2,$$ while Square-DH problem of $U$ and $V$ can be solved.  So by dividing $g^{u^2}$ and $g^{v^2}$, we get $\operatorname{CDH}(U,V)^2$，and at last, compute the square root of $\operatorname{CDH}(U,V)^2$, we get $\operatorname{CDH}(U,V)$

So, can I say that they are equal to each other?
Mostly.

The two problems are actually more closely equivalent in a gap model than in a non-gap model.

Square-DH clearly reduces to CDH either way, but CDH reduces to two calls to Square-DH (you have 3, but you can use $(u-v)^2$ to make it 2).  This is fine if the Square-DH adversary is always right, but maybe the adversary only solves the Square-DH problem with probability $\epsilon$.  The reduction doesn't know whether the adversary succeeded, so it solves CDH with probability only $\epsilon^2$.

If you get to try many times to solve CDH, you have to run the Square-DH adversary $O(2/\epsilon)$ times, and make $O(\epsilon^2)$ guesses, which is somewhat loose... unless you have a DDH oracle, like in the gap model.  In that case, you can filter out wrong Square-DH attempts.  The reduction is still not perfectly tight, though.  It can solve the CDH problem in time $t$ Square-DH calls with a probability which is to first order $t^2\epsilon^2/2$ when $t\ll1/\epsilon$.  So if your adversary is inaccurate and you don't have much time, it's still a loose reduction, but if you have $\approx 1/\epsilon$ time, it becomes tight again.