According to A000001, the number of groups of order $375$ is $7$. How to figure them out?
$5$ groups can be found in this way: Because the number of $125$-order groups is $5$, we can make the direct product of those groups of order $125$ with $Z_3$ and then we get $5$ $375$-order groups.
But how to find the another two groups?
UPD: one of the rest of groups is $C_5^3\rtimes C_3$, and now the task is to find the last missing group.
UPD2: Now I think that a more profound comprehension (to Jyrki Lahtonen's answer) would be appreciated...
It seems to me that the unitriangular group of order $5^3$, GAP ID $(5^3,3)$ has an automorphism of order three.
Let us denote $$ M(a,b,c)=\left(\begin{array}{ccc}1&a&c\\0&1&b\\0&0&1\end{array}\right). $$ Let us consider the mapping $$ \phi:M(a,b,c)\mapsto M(-b,a-b,c-ab+\frac{b(b-1)}2). $$ I brute force checked that $\phi$ is a homomorphism of the group $$U_3(K)=\{M(a,b,c)\mid a,b,c\in K\},$$ $K$ a field such that $1+1\neq0$. Furthermore, $\phi$ has order three.
This gives rise to a non-trivial semidirect product $U_3(K)\rtimes C_3$ with conjugation by a generator of the latter factor corresponding to $\phi$.
With $K=\Bbb{F}_5$ we get the missing seventh group $G$. Its Sylow $5$-subgroup is obviously isomorphic to $U_3(\Bbb{F}_5)$. The only other group of order $375$ with that property in the OP's list is the direct product $H=U_3(\Bbb{F}_5)\times C_3$. But that is not isomorphic to $G$. We easily see that matrices of the form $M(0,0,c)$ form the center of $U_3(K)$, and as $\phi(M(0,0,c))=M(0,0,c)$ this describes the center of $G$ as well. But $Z(H)$ has order $15$.
It is well known and easy to check that $(a,b)\mapsto (-b,a-b)$ is a linear transformation of $K^2$ of order three. The idea was to try and lift that to an automorphism of $U_3(K)$, and $\phi$ popped out. The reason why that might work is that $$U_3(K)/Z(U_3(K))\cong (K^2,+)$$ with the projection given by $M(a,b,c)\mapsto (a,b)$. Furthermore, $M(0,0,1)$ is the commutator of $M(1,0,0)$ and $M(0,1,0)$, so looking for suitable homomorphic images of $M(1,0,0)$ and $M(0,1,0)$ was a natural thing to try. This worked out, but is not really convincing.
I am ashamed to admit that I don't know what the obstructions to lifting an automorphism of $K^2$ to one of $U_3(K)$ are in general. A bit of searching lead me to this excellent post by Jack Schmidt describing the automorphisms of extraspecial $p$-groups (of which $U_3(\Bbb{F}_p)$ are the simplest examples). He also links to articles containing the details.