This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:<p> <i>I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.</i><p> The <b>if</b> part of this statement is correct, while the <b>only if</b> part is not correct.<p> (<b>if</b> holds): <br> Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.<p> (<b>only if</b> fails): <br> The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one $\mathbf{E}$ is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)<p> A more serious reason why <b>only if</b> fails:<br> If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ must be a set rather than a proper class. We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ is a proper class. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.