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This is a cleaner rewrite of my original answer. The answer is no (assuming the surjection is not injective and the smaller set does not have cardinality $1$).

Let $T_A$ be the full transformation monoid on the set $A$. Then the set $C_A$ of constant maps is the unique minimal two-sided ideal of $A$. Since $C_A$ has the same cardinality as $A$, we have $T_B\cong T_A$ if and only if $A$ and $B$ have the same cardinality.

I claim if $|A|\geq 2$, then the unique minimal non-trivial congruence on $T_A$ is to identify all the constant maps to a zero element (absorbing element). Assuming $T_A$ acts on the left of $A$, we have that $T_A$ acts faithfully on the left of $C_A$ by essentially the same action. So any homomorphism that is injective on $C_A$ is injective on $T_A$. On the other hand, if a congruence identifies elements of $C_A$ then the restriction of the congruence to $C_A$ is a system of imprimitivity for the symmetric group $S_A\leq T_A$ acting on the left of $C_A$ which is just the same as its natural action on $A$. This action is $2$-transitive and hence primitive. Thus any non-trivial congruence on $T_A$ must collapse $C_A$.

In conclusion, every proper quotient of $T_A$ has an absorbing element and so can only be a $T_X$ if $|X|=1$. Combined with the fact that $T_A\cong T_B$ iff $A$ and $B$ have the same cardinality, we get the answer is no.

This is a cleaner rewrite of my original answer. The answer is no (assuming the surjection is not injective and the smaller set does not have cardinality $1$).

Let $T_A$ be the full transformation monoid on the set $A$. Then the set $C_A$ of constant maps is the unique minimal two-sided ideal of $T_A$. Since $C_A$ has the same cardinality as $A$, we have $T_B\cong T_A$ if and only if $A$ and $B$ have the same cardinality.

I claim if $|A|\geq 2$, then the unique minimal non-trivial congruence on $T_A$ is to identify all the constant maps to a zero element (absorbing element). Assuming $T_A$ acts on the left of $A$, we have that $T_A$ acts faithfully on the left of $C_A$ by essentially the same action. So any homomorphism that is injective on $C_A$ is injective on $T_A$. On the other hand, if a congruence identifies elements of $C_A$ then the restriction of the congruence to $C_A$ is a system of imprimitivity for the symmetric group $S_A\leq T_A$ acting on the left of $C_A$ which is just the same as its natural action on $A$. This action is $2$-transitive and hence primitive. Thus any non-trivial congruence on $T_A$ must collapse $C_A$.

In conclusion, every proper quotient of $T_A$ has an absorbing element and so can only be a $T_X$ if $|X|=1$. Combined with the fact that $T_A\cong T_B$ iff $A$ and $B$ have the same cardinality, we get the answer is no.

For finite sets the answer is no if $|B|>|A|\geq 2$. Let $T_n$ be the full transformation monoid on $n$-letters (so the monoid of all maps on $n$-letters). Let $C_n$ be the two-sided ideal of constant maps. It is the minimal ideal of $T_n$. Any non-injective homomorphism from $T_n$ must identify all the constant maps. This is because $T_n$ acts faithfully on the left of the constant maps (assuming $T_n$ acts on the left of $1,\ldots, n$) (it looks just like the action of $T_n$ on $1,\ldots n$). So if you identify no constant maps, then you must be injective. If you identify some constant maps you will get a system of imprimitivity for the symmetric group $S_n$ acting on the left of $C_n$ and this action is primitive, being two-transitive. So you must identify all the constant maps. Thus every proper quotient of $T_n$ has a zero element. Since $T_m$ only has a zero element when $m=1$ there is no surjective homomorphism $T_n$ to $T_m$ with $1<m<n$.

This argument shows that the unique minmal congruence on $T_n$ is to identify the constant maps.

Added. I think the same argument should also handle the infinite case. This is a little outside my comfort zone but I think for any set $A$ with at least two elements the symmetric group is two-transitive on $A$ and hence acts primitively and so any congruence on the full transformation monoid on $A$ will have to either be injective on constant maps or collapse them to a point and in the latter case, the image is not a full transformation monoid on more than one element. In the former case, it must be injective on the full transformation monoid. Since the set of constant maps is algebraically determined as the unique minimal two-sided ideal the cardinality of the set $A$ is determined.

This is a cleaner rewrite of my original answer. The answer is no (assuming the surjection is not injective and the smaller set does not have cardinality $1$).

Let $T_A$ be the full transformation monoid on the set $A$. Then the set $C_A$ of constant maps is the unique minimal two-sided ideal of $A$. Since $C_A$ has the same cardinality as $A$, we have $T_B\cong T_A$ if and only if $A$ and $B$ have the same cardinality.

I claim if $|A|\geq 2$, then the unique minimal non-trivial congruence on $T_A$ is to identify all the constant maps to a zero element (absorbing element). Assuming $T_A$ acts on the left of $A$, we have that $T_A$ acts faithfully on the left of $C_A$ by essentially the same action. So any homomorphism that is injective on $C_A$ is injective on $T_A$. On the other hand, if a congruence identifies elements of $C_A$ then the restriction of the congruence to $C_A$ is a system of imprimitivity for the symmetric group $S_A\leq T_A$ acting on the left of $C_A$ which is just the same as its natural action on $A$. This action is $2$-transitive and hence primitive. Thus any non-trivial congruence on $T_A$ must collapse $C_A$.

In conclusion, every proper quotient of $T_A$ has an absorbing element and so can only be a $T_X$ if $|X|=1$. Combined with the fact that $T_A\cong T_B$ iff $A$ and $B$ have the same cardinality, we get the answer is no.

For finite sets the answer is no if $|B|>|A|\geq 2$. Let $T_n$ be the full transformation monoid on $n$-letters (so the monoid of all maps are $n$-letters). Let $C_n$ be the two-sided ideal of constant maps. It is the minimal ideal of $T_n$. Any non-injective homomorphism from $T_n$ must identify all the constant maps. This is because $T_n$ acts faithfully on the left of the constant maps (assuming $T_n$ acts on the left of $1,\ldots, n$) (it looks just like the action of $T_n$ on $1,\ldots n$). So if you identify no constant maps, then you must be injective. If you identify some constant maps you will get a system of imprimitivity for the symmetric group $S_n$ acting on the left of $C_n$ and this action is primitive, being two-transitive. Thus every proper quotient of $T_n$ has a zero element. Since $T_m$ only has a zero element when $m=1$ there is no surjective homomorphism $T_n$ to $T_m$ with $1<m<n$.

This argument shows that the unique minmal congruence on $T_n$ is to identify the constant maps.

For finite sets the answer is no if $|B|>|A|\geq 2$. Let $T_n$ be the full transformation monoid on $n$-letters (so the monoid of all maps on $n$-letters). Let $C_n$ be the two-sided ideal of constant maps. It is the minimal ideal of $T_n$. Any non-injective homomorphism from $T_n$ must identify all the constant maps. This is because $T_n$ acts faithfully on the left of the constant maps (assuming $T_n$ acts on the left of $1,\ldots, n$) (it looks just like the action of $T_n$ on $1,\ldots n$). So if you identify no constant maps, then you must be injective. If you identify some constant maps you will get a system of imprimitivity for the symmetric group $S_n$ acting on the left of $C_n$ and this action is primitive, being two-transitive. So you must identify all the constant maps. Thus every proper quotient of $T_n$ has a zero element. Since $T_m$ only has a zero element when $m=1$ there is no surjective homomorphism $T_n$ to $T_m$ with $1<m<n$.

This argument shows that the unique minmal congruence on $T_n$ is to identify the constant maps.

Added. I think the same argument should also handle the infinite case. This is a little outside my comfort zone but I think for any set $A$ with at least two elements the symmetric group is two-transitive on $A$ and hence acts primitively and so any congruence on the full transformation monoid on $A$ will have to either be injective on constant maps or collapse them to a point and in the latter case, the image is not a full transformation monoid on more than one element. In the former case, it must be injective on the full transformation monoid. Since the set of constant maps is algebraically determined as the unique minimal two-sided ideal the cardinality of the set $A$ is determined.