If we consider a $4\times 4$ complex unitary matrix $V\otimes V\in U(4, \mathbb{C})$, and $V\in M_{2\times 2}(\mathbb{C})$, does this imply that $V$ is also unitary? I am aware that $V\otimes V\in U(4, \mathbb{C})$ if and only if
$$(V\otimes V)(V\otimes V)^{\dagger} = (V\otimes V)(V^{\dagger}\otimes V^{\dagger}) = (VV^{\dagger})\otimes (VV^{\dagger}) = I_{4\times 4}\tag{2}$$
$$(V\otimes V)^{\dagger}(V\otimes V) = (V^{\dagger}\otimes V^{\dagger})(V\otimes V) = (V^{\dagger}V)\otimes (V^{\dagger}V) = I_{4\times 4}\tag{3}$$
Yes.
Lemma. Let $\mathcal{A}$ be some $K$-algebra and $a,b \in \mathcal{A}$ with $a \otimes b = 1$. Then there is some $\lambda \in K^{\times}$ with $a = \lambda \cdot 1$ and $b = \lambda^{-1} \cdot 1$.
Proof: Choose a basis $\{u_i\}$ of $\mathcal{A}$ containing $1$. Write $a = \sum_i \lambda_i u_i$ and $b = \sum_i \mu_i u_i$. Then $$\textstyle 1 \otimes 1 = 1 = a \otimes b = \sum_{i,j} \lambda_i \mu_j u_{i} \otimes u_j.$$ Since the $u_i \otimes u_j$ form a basis, it follows $1 = \lambda_0 \mu_0$ and $0 = \lambda_i \mu_j$ for all $(i,j) \neq (0,0)$. Thus, $\lambda_i = 0$ for all $i > 0$ (take $j=0$) and $\mu_j = 0$ for all $j > 0$ (take $i =0$). Hence, $a = \lambda_0 \cdot 1$ and $b = \lambda_0^{-1} \cdot 1$. $\checkmark$
Lemma. Let $n \geq 1$ and $A \in M_n(\mathbb{C})$. If $A \otimes A \in M_n(\mathbb{C}) \otimes M_n(\mathbb{C})$ is unitary, then $A$ is unitary.
Proof: Since $A \otimes A$ is unitary, we have $A A^{\dagger} \otimes A A^{\dagger} = 1$ in $M_n(\mathbb{C}) \otimes M_n(\mathbb{C})$. By the previous Lemma, $A A^{\dagger} = \lambda \cdot 1$ for some $\lambda \in \mathbb{C}^{\times}$ with $\lambda^2 = 1$. If $\lambda=1$, we are done. The case $A A^{\dagger} = -1$ cannot happen, though, since $A A^{\dagger}$ is positive and hence has positive eigenvalues. $\checkmark$