Let $ A,B $ be matrix groups (with entries in the same field). Then the tensor/Kronecker product $ A \otimes B $ is a matrix group and
$$
\pi: A \times B \to A \otimes B
$$
is a group homomorphism. Taking $ A=B=SU_2 $ we have a map $ \pi: SU_2 \times SU_2 \to SU_4 $ given by
$$
(A,B) \mapsto A \otimes B
$$
The only nontrivial element of the kernel is $ (-1,-1) $. So the image $ SU_2 \otimes SU_2 $ of $ \pi $ is a subgroup of $ SU_4 $ isomorphic to
$$
SU_2 \times SU_2/ (-1,-1) \cong SO_4(\mathbb{R})
$$
Is $ SU_2 \otimes SU_2 $ conjugate to $ SO_4(\mathbb{R}) $ in $ SU_4 $? If so what is a matrix conjugating one to the other? (this part was unanswered for a while but is now answered in the update)
Since matrices in $ SU_2 $ have real trace then all matrices in $ SU_2 \otimes SU_2 $ have real trace. So it is at least plausible that $ SU_2 \otimes SU_2 $ and $ SO_4(\mathbb{R}) $ are conjugate.
Also what is the normalizer in $ SU_4 $ of $ SU_2 \otimes SU_2 $ / the normalizer in $ SU_4 $ of $ SO_4(\mathbb{R}) $? Certainly $ iI $ normalizes $ SO_4(\mathbb{R}) $ and
$$
T=
\zeta_8
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0& 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
normalizes $ SU_2 \otimes SU_2 $.
EDIT: Just to reiterate what Jason DeVito both $ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ are normalized by
$$
T=
\zeta_8
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0& 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
where
$$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0& 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}=SWAP \in O_4(\mathbb{R})
$$
is the $ SWAP $ operator and $ \zeta_8 $ just normalizes the determinant. Both $ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ contain $ -I $ and both do not contain $ iI $. Since
\begin{align*}
(\zeta_8 SWAP)^2&=iI\\
(\zeta_8 SWAP)^4&=-I
\end{align*}
we can combine this with the results from https://arxiv.org/pdf/math/0605784.pdf
to conclude that both $ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ have cyclic $ 4 $ component group generated by $ \zeta_8 SWAP $.
UPDATE: A specific unitary matrix conjugating $ SU_2 \otimes SU_2 $ to $ SO_4(\mathbb{R}) $ is
$$
Q=\frac{1}{\sqrt{2}}
\begin{bmatrix}
1 & 0 & 0 & i \\
0 & i & 1 & 0 \\
0 & i & -1 & 0 \\
1 & 0 & 0 & -i
\end{bmatrix}
$$
in other words there is an isomorphism $ SU_2 \otimes SU_2 \to SO_4(\mathbb{R}) $ given by
$$
M \mapsto Q^{-1} M Q
$$
This is Theorem 1 from https://arxiv.org/pdf/quant-ph/0002045.pdf the matrix $ Q $ is the change of basis from standard basis to the Bell basis well known in quantum computing. This theorem is an interesting and short read for anyone interested in quantum information or just Lie groups. The argument proceeds by concocting a definite form that is preserved by local unitaries $ SU_2 \otimes SU_2 $, instantiating the claims in the answers below that such a form must exist.
Best Answer
Yes, $SU(2)\otimes SU(2)$ is conjugate to the usual $SO(4)$ in $SU(4)$.
There's probably a direct way to see it, but here's some theory. Suppose $G = G_1\times G_2$ is a product of compact Lie groups. Then an irreducible representation (irrep) of $G$ is a tensor product of irreps of $G_1$ and $G_2$. As $G$ is compact, we may assume it acts isometrically, so a rep of $G$ is simply a homomorphism $G\rightarrow U(n)$ for some $n$. If $G$ has no codimension one normal subgroups and it is connected, the $G\rightarrow U(n)$ factors through $SU(n)$: the composition $G\rightarrow U(n)\xrightarrow{\det} S^1$ has connected image, so is either trivial or surjective. If surjetive, the kernel is a codim 1 normal subgroup.
Now, let's specialize to the case $G = SU(2)\times SU(2)$. We are looking for $4$-dim reps of $G$. Such a rep is a sum of irreps. The irreps of $SU(2)$ are well known: there is a a unique irrep of each dimension $n\geq 1$. I'll use the notation $(n\times m)$ to refer to the tensor product of the $n$-dim irrep of $SU(2)$ with the $m$-dim irrep of $SU(2)$.
Then the possible $4$-dim reps of $SU(2)\times SU(2)$ are (up to permuting the factors):
Now, the reps given by $SU(2)\otimes SU(2)$ and $SO(4)$, when thought of as reps of $SU(2)\times SU(2)$, both have kernel precisely $\pm(1,1)$. In the above list, apart from entries $5$ and $7$, the kernel contains an entire factor of $SU(2)$. Entry $5$ is injective.
So, $SU(2)\otimes SU(2)$ and $SO(4)$ are isomorphic representations. By definition, then, there is a matrix $A\in Gl(4,\mathbb{C})$ where $A$ conjugates $SU(2)\otimes SU(2)$ to $SO(4)$. But can we take $A$ to be in $SU(4)$?
Yes, and here's how. First, we know that for any $k\in SO(4)$, that $$(A^{-1}ka)(\overline{A^{-1}kA})^t = I$$ since $ASO(4)A^{-1}= SU(2)\otimes SU(2) \subseteq SU(4)$. This equation can be rearranged to read $ k A\overline{A}^t \overline{k}^t = A\overline{A}^t$, which means that the $SO(4)$ action preserves the Hermiatian forms represented by $I$ and by $A\overline{A}^t$. But the $SO(4)$ rep is irreducible, so from this MSE question, we must have $A\overline{A}^t = \lambda I$ for some non-zero $\lambda \in \mathbb{C}$. As the diagonal entries of $A\overline{A}^t$ are of the form $\lambda =\sum_j |a_{ij}|^2\geq 0$, we conclude $\lambda$ must be a positive number. Then the matrix $B:=\frac{1}{\sqrt{\lambda}}A$ is in $U(4)$ and conjugates $SU(2)\otimes SU(2)$ to $SO(4)$. Finally, to find an example in $SU(4)$, use $C:= \overline{\mu} B$ where $\mu$ is any $4$th root of $\det(B)$.
I'm not sure what the normalizer is. I'll keep working on it and update this answer if I figure it out.